a potpourri of algebra

A Potpourri of Algebra

Khor Shi-Jie

March 27, 2012

1

Experience has taught me that there are two distinct phases in the grasping of Mathematical Olympiad.First and foremost, one has to understand the content which is being taught during MathematicalOlympiad lessons or through reference materials. This includes all the relevant theorems of the fourmajor topics in MO: Algebra, Geometry, Combinatorics and Number Theory. Achieving this first stepshould not take the student too much time as it does not take much effort to learn such knowledge that isreadily available to general students of mathematics. Of much higher priority is the second phase wherebystudents internalise the usage and application of these learned knowledge. One must understand and ap-preciate various creative methods of applying learnt theorems in order to excel in MO competitions.These notes aim to facilitate your second preparatory step in mastering the Mathematical Olympiad.

Algebra is a major topic in Junior Section, be it in the first round or the second round. Recentlyparticipants have complained that questions in the SMO Junior Section are becoming more challengingand ”unapproachable”. This is mainly due to the fact that the problems which are appearing in recentSMO papers are fresh and some students have no experience of dealing with such questions. Nonetheless,a strong foundation in algebra plus the application of suitable strategies will help you solve these problems.Note my usage of the term ”strategies” instead of ”theorems”. A good MO student usually thinks interms of strategies to tackle the problem instead of theorems used to solve the problem. I have organisedpast year SMO problems as well as problems from other countries according to the strategies used tosolve the problems. Hopefully this will improve your manipulation skills in solving algebra problems.

Contents

1 Substitution 31.1 Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Completing the Square 112.1 Simplification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Proving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Factorisation 173.1 Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.2 Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.3 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Miscellaneous Techniques in Algebra 244.1 Discriminant and Vieta’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.2 Method of differences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.3 Method of fixed ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.4 Geometric constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2

Chapter 1

Substitution

The aim of using substitution (in junior section) is to simplify an expression which is otherwise too com-plicated or untidy to study carefully. Granted, it is possible to solve such problems without substitution,but it will take much more effort and time in competitions.

Here are several types of questions which can be solved using careful substitution:

1.1 Evaluation

In such questions, students are often given an expression with very large numbers and they are asked tosimplify or evaluate such expressions. Take a look at SMO(J)2008 First Round P23:

Evaluate(20202 − 20100)(201002 − 1002)(20002 + 20100)

20106 − 106

This is a nasty expression to work with directly without any tools, especially during a competition. Weare going to substitute the number 2010 in the equation using a variable to simplify our calculation. Thesubstitution x = 2010 sticks out like a sore thumb. So let us see if it works...

[(x+ 10)2 − 10x][(10x)2 − 1002][(x− 10)2 + 10x]

x6 − 106

It looks more manageable now and can possibly be solved algebraically from now onwards, but why stophere? Let us substitute y = 10 to further simplify the equation...

[(x+ y)2 − xy][(xy)2 − y4][(x− y)2 + xy]

x6 − y6

From this expression, we derive that the numerator can be simplified into:

[(x+ y)2 − xy][(xy)2 − y4][(x− y)2 + xy] = (x2 + xy + y2)y2(x+ y)(x− y)(x2 − xy + y2)

= y2(x3 + y3)(x3 − y3)

= y2(x6 − y6)

Bingo! By using substitution we can easily see that the expression factorises and recombines to form asum of cube and difference of cube expression respectively. It cancels out the denominator and producesthe elegant answer of 100.

Here’s another example which is very similar to a problem that students have encountered in the MOPclass.

3

CHAPTER 1. SUBSTITUTION 4

Given that x =3 +√

5

2, evaluate

2a5 − 5a4 + 2a3 − 8a2

a2 + 1

Before substitution, we have to modify the condition given to make it look more appetising...

x =3 +√

5

2⇔ 2x+ 3 =

√5⇔ x2 − 3x+ 1 = 0⇔ x2 = 3x− 1

Upon multiple substitution of x2 into the equation, one should obtain the answer of -1.

1.2 Comparison

It is useful for one to substitute a large number or an expression when asked to compare the size of twocomplicated expressions. For example:

Compare the size of A =5678901234

6789012345and B =

5678901235

6789012347

To solve this problem, we let x be the value 5678901234 and y be the value 6789012345. We see that

A =x

ywhile B =

x+ 1

y + 2. Taking the difference A−B, we have:

x

y− x+ 1

y + 2=

2x− yy(y + 2)

which is obviously larger than 0 since 2x > y and y > 0. Hence we have A > B.

1.3 Solving Equations

The method of substitution is often used to solve equations of higher degree, exponential equations,trigonometric equations, logarithmic equations, etc. Here are several examples:

Solve (6x+ 7)2(3x+ 4)(x+ 1) = 6

I hope that the word “substitution” comes across your mind automatically when you see this prob-lem. Technically, it is possible to solve this problem by expansion and using the rational root theorem.However, this takes plenty amount of time and we should instead try to think of methods to make eachbracket look as similar as possible. This is possible by multiplying 12 to both sides of the equation,obtaining

(6x+ 7)2(6x+ 8)(6x+ 6) = (6x+ 7)2[(6x+ 7)2 − 1] = 72

Upon seeing this expression, the substitution y = (6x+ 7)2 should spring into your mind, simplifying the

equation into a quadratic equation with roots 9 and -8. We finally obtain x1 = −2

3or x2 = −5

3.

Let us try a problem which involves exponents:

Solve 24x + (2x − 2)4 − 34 = 0

Obviously expanding (2x − 2)4 is not wise. Again, one should think of substitution intuitively uponseeing exponents in an equation. The trickier question is to choose the right value to substitute.

Do you gain by using the substitution y = 2x? Upon substitution we obtain y4 + (y − 2)4 − 34 = 0.Once you expand the (y − 2)4 term, you have to deal with a quartic equation and that can be quite


troublesome. Similarly, the substitution y = 2x − 2 does not help.

What if we take the value in between, say y = 2x−1? Interestingly, the equation (y+1)4+(y−1)4−34 = 0is considerably tamer because the alternate terms in the expansion cancels out each other. After ex-panding the terms (which is simple with the aid of binomial theorem or Pascal’s Triangle), we gety4 + 6t2 − 16 = 0, which can be solved as if it is a quadratic equation. This gives us the solutionsy =

√2 and y = −

√2. We reject the latter solution since y must be positive. Finally, we solve that

x = log2

√2 + 1.

As a side note, the theme of (x + y)k + (x − y)k is worth remembering and occurs in other kinds ofproblem too. Try finding the integer closest to (2 +

√2)6.

Finally, let us look at this mean-looking question: Evaluate

1 +1

1 +1

1 +1

1 + · · ·

To solve this, let x be the expression above. Note that x is exactly the expression at the denominator of

the fraction. We then have x = 1 +1

x, which can be solved easily as a quadratic equation. Upon solving,

the expression is equal to1 +√

5

2. (In case you haven’t realise that’s the golden ratio)

1.4 Factorisation

Substitution helps in factorisation too, provided if you choose the right expression to substitute.

Factorise 4(x+ 5)(x+ 6)(x+ 10)(x+ 12)− 3x2

Do I need to expand the entire expression? Sadly, the answer is yes. However, through substitutionwe can greatly simplify the process. Simply note that 5 × 12 = 6 × 10. Upon grouping useful termstogether and expanding, we obtain

4(x+ 5)(x+ 6)(x+ 10)(x+ 12)− 3x2 = 4(x+ 5)(x+ 12)(x+ 6)(x+ 10)− 3x2

= 4(x2 + 16x+ 60)(x2 + 17x+ 60)− 3x2

Now, y = x2 + 16x+ 60 looks like a smart substitution. Applying that to our equation, we obtain

LHS = 4y(y + x)− 3x2

= 4y2 + 4xy − 3x2

= (2y − x)(2y + 3x)

= (2x2 + 31x+ 120)(2x+ 35x+ 120)

= (2x+ 15)(x+ 8)(2x2 + 35x+ 120)

If you are an avid reader of my blog (hcmop.wordpress.com), you probably have heard about symmetricpolynomials and elementary symmetric polynomials (ESP). Here’s an application of ESP to factorise thefollowing polynomial:

Factorise x3 + y3 − 3xy + 1

Introduce the notations σ1 and σ2 to indicate the ESPs for second degree polynomials x + y and xy


respectively. We derive:

x3 + y3 − 3xy + 1 = σ31 − 3xy2 − 3x2y − 3xy + 1

= σ31 + 1− 3σ1σ2 − 3σ2

= (σ1 + 1)(σ21 − σ1 + 1)− 3σ2(σ1 + 1)

= (σ1 + 1)(σ21 − σ1 − 3σ2 + 1)

= (x+ y + 1)(x2 + y2 − xy − x− y + 1)

Do read my blog for a more detailed discussion on symmetric polynomials and ESPs.

1.5 Problem Set

1. (SMO(J)2007P35) Find the largest integer N such that both N + 496 and N + 224 are perfectsquares.

2. (AIME05P7) Let x =4

(√

5 + 1)( 4√

5 + 1)( 8√

5 + 1)( 16√

5 + 1). Find (x+ 1)48.

3. (SMO(S)2011P6) Determine the value of2

1√2 + 4√

8 + 2+

1√2− 4√

8 + 2

.

4. (SMO(J)2006P25) What is the product of the real roots of the equation

x2 + 90x+ 2027

3=√x2 + 90x+ 2055

5. (SMO(J)2006P34) Suppose that the two roots of the equation

1

x2 − 10x− 29+

1

x2 − 10x− 45− 2

x2 − 10x− 69= 0

are α and β. Find α+ β

6. (SMO(S)2006P10) Let a and b be positive real numbers such that

1

a− 1

b− 1

a+ b= 0

Find the value of (a

b+b

a)2

7. Suppose t =a2

a2 + 2b2+

b2

2a2 + b2. Find the minimum value of t.

8. Given that a, x, y are real numbers that satisfy

log2a x+ log2

a y − loga (xy)2 ≤ 2, loga y ≥ 1

Find the range of loga x2y.

9. (SMO(O)96P3) Let a > 1 be an integer. Find all integers x such that

(

√a+

√a2 − 1)x + (

√a−

√a2 − 1)x = 2a

10. Solve the following the system of equations where x, y, z are positive numbers:

xxyz = y2

yxyz+1 = z3

zxyz+2 = x4


11. Solve the equation x4 − 6x3 + 11x2 − 6x+ 1 = 0

12. (USAMO78P1) Given that a, b, c, d, e are real numbers such that

a+ b+ c+ d+ e = 8

a2 + b2 + c2 + d2 + e2 = 16

Determine the maximum value of e.


1.6 Solutions

1. There are 2 ways to solve this problem. Firstly, we can let a2 = N + 496 and b2 = N + 224.We derive that a2 − b2 = 272 ⇒ (a − b)(a + b) = 24 × 17. Hence, the possible solutions for(a − b, a + b) = (1, 272), (2, 136), (4, 68), (8, 34), (16, 17). Upon solving the equations, we only have3 sets of integer solutions (a, b) = (69, 67), (36, 32), (21, 13). The largest N occurs when a is 69.Hence we obtain N = 4265.

Another way to solve it is to note that the difference between two squares is an even number.To maximise N , the difference a − b must be as small as possible. N will take maximum valuewhen a− b is 2. Hence we can construct the equation a2 − (a− 2)2 = 272 and obtain the solutiona = 69, N = 4265.

2. Substitute k = 16√

5 into the equation, we obtain

x =4

(x8 + 1)(x4 + 1)(x2 + 1)(x+ 1)

=4

1 + x+ x2 + · · ·+ x15

=4

x16 − 1

x− 1

=4(x− 1)

x16 − 1

=4( 16√

5− 1)

5− 1

=16√

5− 1

So we have (x+ 1)48 = 125.

3. Let x = 4√

2. We have

21√

2 + 4√

8 + 2+

1√2− 4√

8 + 2

=2

1

2 + x3 + x2+

1

2− x3 + x2

=2

2(x2 + 2)

(x2 + 2)2 − x6

=(x2 + 2)2 − x6

x2 + 2

=(√

2 + 2)2 − 2√

2]√2 + 2

=6 + 2

√2√

2 + 2

= 4−√

2

4. Let y = x2 +90x+2027. We havey

3=√y + 28. Solving yields y = 21 or y = −12. When y = −12,

there is no solution for x. When y = 21, we have x2 + 90x + 2006 = 0 and hence the product ofroots is 2006.

5. Let y = x2 − 10x− 45. We have1

y + 16+

1

y− 2

y − 24= 0. Clearing the denominators, we obtain

y = −6. Hence we have x2 − 10x− 39 = 0 which suggests that the sum of roots is 10.


6. Let x =a

b. We have a = bx. Through substitution, we obtain:

1

a− 1

b− 1

a+ b=

1

bx− 1

b− 1

bx+ b

=1

x− 1− 1

1 + x

= 0

Solving the above equation, we obtain x =

√5− 1

2. This gives us (

a

b+b

a)2 = (x+

1

x)2 = 5.

7. Let x = a2 + 2b2, y = 2a2 + b2, we have

t =y − x+ y

3x

+x− x+ y

3y

=2y − x

3x+

2x− y3y

=2y

3x+

2x

3y− 2

3

By using AM-GM inequality or the fact that m + n ≥ 2√mn, we have

2y

3x+

2x

3y≥ 4

3. Hence, the

minimum value of t is2

3.

Using a variable to substitute the denominator is a useful technique since this allows you to simplifyterms very easily.

8. Let m = loga x and n = loga y. We have m2 + n2 − 2(m+ n) ≤ 2 and n ≥ 1. Upon completing thesquare, the first equation becomes (m − 1)2 + (n − 1)2 ≤ 4. Together with the second constraint,it forms a semicircle with radius 2 and centre at (1, 1).

We shall employ the method of linear programming to find the range of 2m+n. Let k = 2m+n⇔n = −2m+ k. We construct the lines y = −2x+ k such that k takes the maximum and minimumvalue in the graph respectively at permitted values of (x, y) as bounded by the semicircle.

All is left for us to do is to determine the y-intercept of the two linear functions. With the aid ofgeometry we can derive that k ∈ [−1, 3 + 2

√5].

9. Let y =√a+√a2 − 1. We also have

1

y=

1√a+√a2 − 1

=√a−√a2 − 1. This gives us

yx +1

yx= 2a y2x − 2ayx + 1 = 0


By using the quadratic formula, we obtain yx = (√a+√a2 − 1)x = a ±

√a2 − 1. Hence the

possible values for x are 2 and -2.

10. From the first equation, we have y = xxyz2 . Substitute into the second equation, we have z =

x(xyz)(xyz+1)

6 Substituting for the final time in the third equation, we have x(xyz)(xyz+1)(xyz+2)

6 = x4.

Let a = xyz. We either have x = 1 or a(a + 1)(a + 2) = 24. For the first case, there are twopossible solutions (1, 1, 1) and (1,−1, 1). For the second case, upon rearrangement and factorisa-tion we obtain (a − 2)(a2 + 5a + 12) = 0. Hence, the only solution is xyz = 2. Examining thesystem of equations again, we see that we either have x = y = z or x = −y = −z. This gives usthe two other solutions ( 3

√2, 3√

2, 3√

2) and ( 3√

2,− 3√

2,− 3√

2).

11. Note that x = 0 is not a solution. Dividing the equation by x2 throughout, we have x2− 6x+ 11−6

x+

1

x2= 0. Use the substitution y = x+

1

x. We have y2−6x+ 9 = 0 which gives us y = 3. Hence,

x+1

x= 3 and x =

3±√

5

2.

12. From a + b + c + d = 8 − e, we get the inspiration to use the substitution a =8− e

4+ α, b =

8− e4

+ β, c =8− e

4+ γ, d =

8− e4

+ δ, where α+ β + γ + δ = 0. We then have

a2 + b2 + c2 + d2 = 4(8− e

4)2 + 2(α+ β + γ + δ)(

8− e4

) + α2 + β2 + γ2 + δ2

= 4(8− e

4)2 + α2 + β2 + γ2 + δ2

≥ (8− e)2

4

We also have a2 + b2 + c2 + d2 = 16− e2 and hence 16− e2 ≥ (8− e)2

4. Upon solving this quadratic

inequality, we have 0 ≤ e ≤ 16

5. Hence the maximum value of e is

16

5.

This proof is unique because it does not utilise any inequality theorems at all. Mean substitu-tion can be useful when the sum of several variables are given.

Chapter 2

Completing the Square

Completing the square is another useful technique in simplifying expressions for closer inspection. It iscommonly applied in SMO problems when a high degree polynomial is given. This techniques is usuallyused in simplification and finding extremal values.

This techniques is not restricted to expansion of squares only. Sometimes, one has to complete thecube or expressions with higher degree. Some notable formulas include:

1. (a+ b)2 = a2 + 2ab+ b2

2. (a+ b+ c)2 = a2 + b2 + c2 + 2ab+ 2bc+ 2ca

3. (a+ b)3 = a3 + 3a2b+ 3ab2 + b3

2.1 Simplification

It is often useful to complete the squares in problems which involves variables of higher degree or expres-sions which involves surds. For example,

Simplify√x− 1 + 2

√x− 2 +

√x− 1− 2

√x− 2

The key in this problem is to realise the fact that x − 1 is equal to x − 2 + 1. The original expres-sion can be written as:√

x− 1 + 2√x− 2 +

√x− 1− 2

√x− 2 =

√(x− 2) + 2

√x− 2 + 1 +

√(x− 2)− 2

√x− 2 + 1

=

√(√x− 2 + 1)2 +

√(√x− 2− 1)2

=√x− 2 + 1 + |

√x− 2− 1|

Hence the orginal expression is equal to 2 when 2 ≤ x < 3 and 2√x− 2 when x ≥ 3


We know that quadratic equations can be solved using the technique of completing the squares. In fact,the quadratic formula which produces the roots of a quadratic equation is derived using the technique ofcompleting the squares.

The technique of completing the squares is very useful to check if an equation has any solutions atall. For example,

Prove that x4 + 3x2 + 2x+ 6 = 0 has no real roots

To prove this, we have to complete the square in the following form:

x4 + 3x2 + 3x+ 6 = (x2 + 1)2 + (x+ 1)2 + 4 = 0

Since each term in the expression is always strictly greater than zero, there is no solution to the equation.

11

CHAPTER 2. COMPLETING THE SQUARE 12

2.3 Proving Inequalities

Probably the most useful application of completing the squares is in proving inequalities. In juniorsection, students are not expected to have learned advanced inequalities such as AM-GM inequalityor Cauchy-Schwarz inequality. Students have to rely on basic algebraic manipulation skills in order toprove inequalities in competitions. The technique of completing the squares is very handy since all squaresmust be greater or equal to zero. Let us take a look at the following problem proposed by Titu Andreescu:

Let a, b, c be real numbers. Prove that the numbers a − b2, b − c2, c − d2, d − a2 cannot be all larger

than1

4.

The solution comes intuitively if you are used to solving inequalities using the method of completing

the squares. Let us suppose that it is possible for all for expressions to be larger than1

4at the same

time, i.e.

a− b2 > 1

4, b− c2 > 1

4, c− d2 > 1

4, d− a2 > 1

4

By adding the four expressions above together, we obtain

a+ b+ c+ d− (a2 + b2 + c2 + d2) < 1

Moving all terms to the right and completing the squares, we have

(1

2− a)2 + (

1

2− b)2 + (

1

2− c)2 + (

1

2− d)2 < 0

which is obviously a contradiction.

Students in junior section should take note of the two fundamental ways to prove inequalities, whichare the method of difference and the method of division respectively. To prove that A ≥ B, one canattempt to prove that

1. A−B ≥ 0 or

2.A

B≥ 1.

2.4 Problem Set

1. (SMO(S)2011 P3 First Round) Let x be a real number. If a = 2011x + 9997, b = 2011x + 9998,c = 2011x+ 9999, find the value of a2 + b2 + c2 − ab− bc− ac.

2. (SMO(S)2011 P5 First Round) Suppose x, y are real numbers such that1

x− 1

2y=

1

2x+ y. Find

the value ofx2

y2+y2

x2.

3. (SMO(J)2011 P19 First Round) Let a, b, c, d be real numbers such that

a2 + b2 + 2a− 4b+ 4 = 0

c2 + d2 − 4c+ 4d+ 4 = 0

Let m and M be the minimum and maximum value of (a− c)2 + (b−d)2 respectively. Find m×M .

4. Suppose (x− z)2 − 4(x− y)(y − z) = 0. Prove that y is the mean of x and z.

5. Suppose x, y, z are distinct real numbers. Prove that (1

y − z)2 + (

1

z − x)2 + (

1

x− y)2 = (

1

y − z+

1

z − x+

1

x− y)2.

6. Suppose a, b, c, d are positive real numbers that satisfy a4 + b4 + c4 + d4 = 4abcd. Prove thata = b = c = d.


7. (T. Andreescu) Find all real solutions to the system of equations

x+ y =√

4z − 1

y + z =√

4x− 1

z + x =√

4y − 1

8. (IMO Longlist 1970 P37) Solve the set of simultaneous equations

v2 + w2 + x2 + y2 = 6− 2z

v2 + w2 + x2 + z2 = 6− 2y

v2 + w2 + y2 + z2 = 6− 2x

v2 + x2 + y2 + z2 = 6− 2w

w2 + x2 + y2 + z2 = 6− 2v

9. (T. Andreescu) Find all real triplets (x, y, z) that satisfy x4 + y4 + z4 − 4xyz = −1

10. Solve the system of equations

x3 − 9(y2 − 3y + 3) = 0

y3 − 9(z2 − 3z + 3) = 0

z3 − 9(x2 − 3x+ 3) = 0

11. (SMO(J)2011 Second Round P1) Suppose a, b, c, d > 0 and x =√a2 + b2, y =

√c2 + d2. Prove that

xy ≥ ac+ bd.

12. Suppose x, y, z are real numbers that satisfy xy + yz + xz = −1. Prove that x2 + 5y2 + 8z2 ≥ 4

13. Suppose a, b, c are positive real numbers. Prove that for any real numbers x, y, z, we have

x2 + y2 + z2 ≥ 2

√abc

(a+ b)(b+ c)(c+ a)(

√a+ b

cxy +

√b+ c

ayz +

√c+ a

bzx)


2.5 Solutions

1. Note that a2 + b2 + c2 − ab − bc − ca =1

2[(a − b)2 + (b − c)2 + (c − a)2]. By plugging in all the

expressions in the original question, we obtain the answer 3.

2. We note thatx2

y2+y2

x2= (

y

x− x

y)2 + 2. From the condition given in the question, we have

1

x− 1

2y=

1

2x+ y⇔ 2x+ y

x− 2x+ y

2y= 1

⇔ y

x− x

y= −1

2

Hencex2

y2+y2

x2=

9

4

3. By completing the squares for the two equations given, we obtain

(a+ 1)2 + (b− 2)2 = 1

(c− 2)2 + (d+ 2)2 = 4

which are equations of circles with radius 1 at (-1, 2) and radius 2 at (2,-2) respectively. Thequantity (a− c)2 + (b− d)2 represents the distance between a point in the first circle and anotherpoint in the second circle. Since the distance between the centres of the circles is 5 (by PythagorasTheorem), the maximum and minimum value of (a− c)2 + (b− d)2 are 8 and 2 respectively. Hencethe solution is 16.

4. By expanding the entire expression, we have

x2 − 2xz + z2 − 4xy + 4y2 + 4xz − 4yz = 0

Upon rearrangement, we obtain

(x+ z)2 − 4(x+ z)y + 4y2 = 0

By completing the square, we have[(x+ z)− 2y]2 = 0

It follows that y =x+ z

2.

5. From the fact that (x − y) + (y − z) + (z − x) = 0, we have1

(x− y)(y − z)+

1

(y − z)(z − x)+

1

(z − x)(x− y)= 0. Hence,

(1

y − z)2 + (

1

z − x)2 + (

1

x− y)2 = (

1

y − z+

1

z − x)2 − 2

1

y − z· 1

z − x+ (

1

x− y)2

= (1

y − z+

1

z − x)2 + 2(

1

y − z+

1

z − x) · 1

x− y+ (

1

x− y)2

= (1

y − z+

1

z − x+

1

x− y)2

6. We shall attempt to move all terms to left hand side, complete the squares and use the propertythat no square is negative to solve this question. Since a4 + b4 + c4 + d4 = 4abcd, we have2a4 + 2b4 + 2c4 + 2d4 = 8abcd. Moving all the terms to the left, we complete the squares inthe following manner:

2a4 + 2b4 + 2c4 + 2d4 − 8abcd =(a2 − b2)2 + 2a2b2 + (b2 − c2)2 + 2b2c2 + (c2 − d2)2 + 2c2d2

+ (d2 − a2) + 2a2d2 − 8abcd

=(a2 − b2)2 + (b2 − c2)2 + (c2 − d2)2 + (d2 − a2)2 + 2(ab− cd)2

+ 2(bc− da)2

=0


This gives us the following system of equations:

a2 − b2 = 0

b2 − c2 = 0

c2 − d2 = 0

d2 − a2 = 0

ab− cd = 0

bc− da = 0

Since a, b, c, d are positive real numbers we must have a = b = c = d.

7. We start by adding the three equations together and try to complete the squares. By taking thesum and moving all terms to the left hand side of the equation, we have

2x+ 2y + 2z −√

4x− 1−√

4y − 1−√

4z − 1 = 0

Next, we divide the equation by 2 so that the coefficient of the variables in the surds become 1.

x+ y + z −√x− 1

4−√y − 1

4−√z − 1

4= 0

Finally, we note that x− 1

4−√x− 1

4+

1

4= (

√x− 1

4− 1

2)2. By completing the squares, we have

(

√x− 1

4− 1

2)2 + (

√y − 1

4− 1

2)2 + (

√z − 1

4− 1

2)2 = 0

Since no square is negative, it follows that x = y = z =1

2is the only set of solution.

8. Without loss of generality, we shall discuss this problem in three different cases:

(a) z is not equal to any of the four other variables

(b) y = z and z is not equal to the three other variables

(c) All variables are equal

For the first case, we find the difference between the first equation and the other four equations.We have

y2 − z2 = 2y − 2z ⇔ y = 2− zx2 − z2 = 2x− 2z ⇔ x = 2− zw2 − z2 = 2w − 2⇔ w = 2− zv2 − z2 = 2v − 2z ⇔ v = 2− z

which implies that v = w = x = y = 2 − z. Substitute back into the first equation, we have4(2− z)2 = 6− 2z which gives us the solutions (1,1,1,1,1) and (-1/2,-1/2,-1/2,-1/2,5/2).For the second case, we also have the following relationships:

x2 − z2 = 2x− 2z ⇔ x = 2− zw2 − z2 = 2w − 2⇔ w = 2− zv2 − z2 = 2v − 2z ⇔ v = 2− z

Substitute these quantities back into the first equation , we have 3(2−z)2 +z2 = 6−2z which givesus the solutions (1,1,1,1,1) and (1/2,1/2,1/2,3/2,3/2).

Finally when all of the variables are the same, we can add all the equations and complete the

square, hence obtaining (2z +1

2)2 =

125

4which gives us the solutions (1,1,1,1,1) and (-3/2,-3/2,-

3/2,-3/2,-3/2). Hence, the final solutions are (1,1,1,1,1), (-3/2,-3/2,-3/2,-3/2,-3/2) and permuta-tions of (-1/2,-1/2,-1/2,-1/2,5/2) and (1/2,1/2,1/2,3/2,3/2).

P/S: Included this question initally because I thought this question can only be solved throughcompleting the square. There’s no need for that actually.


9. By completing the square for the first 2 terms, we obtain:

(x2 − y2) + 2x2y2 + z4 − 4xyz = −1

By subtracting 2z2 from the expression, we can complete the square another time:

(x2 − y2)2 + (z2 − 1)2 + 2x2y2 + 2z2 − 4xyz = 0

Completing the square for one last time, we obtain:

(x2 − y2)2 + (z2 − 1)2 + 2(xy − z)2 = 0

Hence, the solutions are (1, 1, 1), (−1, 1,−1), (1,−1,−1).

10. We can complete the cube and rewrite the system of equations as

(y − 3)3 = y3 − x3

(z − 3)3 = z3 − y3

(x− 3)3 = x3 − z3

Upon addition we obtain (x− 3)3 + (y − 3)3 + (z − 3)3 = 0. Without loss of generality, we assumethat x ≥ 3. From the equation z3 − 9(x2 − 3x + 3) = 0, we note that z3 − 27 = 9x(x − 3) andhence z ≥ 3. Similarly we have y ≥ 3. This implies that the only solution to the equation isx = y = z = 3.

11. We shall prove that x2y2 ≥ (ab+ cd)2 since all numbers are non-negative. We have

x2y2 = (a2 + b2)(c2 + d2)

= a2c2 + b2d2 + a2d2 + b2c2

≥ a2c2 + b2d2 + 2abcd

= (ab+ cd)2

due to the fact that a2d2 + b2c2 − 2abcd = (ad− bc)2 ≥ 0.

We can also prove this question using properties of discriminant. Consider the quadratic equation(a2+b2)x2+2(ac+bd)x+c2+d2 = 0. Upon completing the squares, we obtain (ax+c)2+(bx+d)2 = 0.Since the left hand side of the equation is greater or equals to zero, the discriminant of the quadraticequation must be smaller or equal to zero, i.e.:

4(ac+ bd)2 − 4(a2 + b2)(c2 + d2) ≤ 0⇔ xy ≥ ac+ bd

12. By shifting all terms to the left hand side of the inequality, we have:

x2 + 5y2 + 8z2 − 4 = x2 + 5y2 + 8z2 + 4(xy + yz + zx)

= (x+ 2y + 2z)2 + (y − 2z)2 ≥ 0

13. Shifting all terms to the left, we have

x2 + y2 + z2 − 2

√abc

(a+ b)(b+ c)(c+ a)(

√a+ b

cxy +

√b+ c

ayz +

√c+ a

bzx)

=

[b

b+ cx2 +

a

a+ cy2 − 2

√ab

(b+ c)(c+ a)xy

]+

[c

c+ ay2 +

b

a+ bz2 − 2

√bc

(c+ a)(a+ b)yz

]

+

[c

b+ cx2 +

a

a+ bz2 − 2

√ca

(b+ c)(a+ b)xz

]

=ab

[x√

a(b+ c− y√

b(c+ a)

]2+ bc

[y√

b(c+ a− z√

c(a+ b)

]2+ ca

[z√

c(a+ b− x√

a(b+ c)

]2≥0

Hence the original inequality stands.

Chapter 3

Factorisation

Factorisation is a useful technique that is widely employed in MO. There are many things that we cando with a factorised form of an expression. Here are some of the factorisation formula that we oftenencounter:

1. ab+ ac+ bd+ cd = (a+ d)(b+ c)

2. ax2 − (α+ β)x+ αβ = a(x− α)(x− β)

3. a2 − b2 = (a+ b)(a− c)

4. a3 + b3 = (a+ b)(a2 − ab+ b2

5. a3 − b3 = ((a− b)(a2 + ab+ b2)

You may find the following factorisation formulas helpful too:

1. a3 + b3 + c3 − 3abc =1

2(a+ b+ c)[(a− b)2 + (b− c)2 + (c− a)2].

2. a4 + 4b4 = (a2 + 2ab+ 2b2)(a2 − 2ab+ 2b2) (Sophie Germain Identity)

3. ak − bk = (a− b)(ak−1 + ak−2b+ ak−3b2 + · · ·+ bk−1) for positive integer k

4. ak + bk = (a+ b)(ak−1 − ak−2b+ ak−3b2 − · · ·+ bk − 1) for odd integer k

However, the factorisation of expressions in most olympiad problems are not obvious and require tech-niques beyond these common formulas. These techniques include:

1. Substitution

2. Coefficient determination method

3. Factor theorem and rational root theorem

I have already highlighted the use of substitution to factorize expression in the first chapter. On the otherhand, the coefficient determination method is often employed when the factors have more than two terms.For example, to factorise 2x2 = 7xy−4y2−3x+6y−2, we expand the expression (2x−y+a)(x+4y+ b)and compare the coefficients to solve for a and b.

Another very powerful tool which is often employed to factorise symmetric or cyclic polynomial is factortheorem. The factor theorem simply states that if P (a) = 0 for some polynomial P (x), then (x− a) is afactor of the polynomial. Let us use this theorem to factorise a3(b− c) + b3(c− a) + c3(a− b). By takinga as the principle variable and evaluating P (b), we have:

P (b) = b3(b− c) + b3(c− b) + c3(b− b) = 0

Hence (a− b) is a factor of the expression. Similarly, (b− c) and (c−a) are also factors of the expression.Now we try to let a = −b− c and see what we obtain:

P (−b− c) = (−b− c)3(b− c) + b3[c− (−b− c)] + c3[(−b− c)− b] = 0

17

CHAPTER 3. FACTORISATION 18

This suggests that (a + b + c) is a factor too. It seems like the expression is in the form of A(a + b +c)(a − b)(b − c)(c − a) where A is an unknown constant. Upon verifying with the initial expression, weobtain A = 1 and hence a3(b− c) + b3(c− a) + c3(a− b) = (a+ b+ c)(a− b)(b− c)(c− a).

It is often useful to check values such as a = −b − c to see if (a + b + c) is a factor of the expres-sion using factor theorem (most of the time, it is). In addition, if the expression is cyclic, note that thefactors of the cyclic expression will be cyclic too. So if (a−b) is a factor of the cyclic expression, naturally(b− c) and (c− a) are also factors of the expression.

Finally, the rational root theorem is a tool that enables one to find the factors of single variable polyno-

mials. The rational root theorem states that if x =p

qis a root of a polynomial P (x) = anx

n+ · · ·+a0 and

(p, q) = 1, then we must have p|a0 and q|an. For example, suppose we want to factorise x3−2x2−2x−3.By the rational root theorem, the only possible rational roots are 1,−1, 3,−3. Upon substituting backinto the equation, we realise that x = 3 is a root of the polynomial and hence we can factorise thepolynomial into (x− 3)(x2 + x+ 1).


Well, I guess you have realised that this is the application that almost all techniques in algebra have incommon. The usefulness of factorisation technique is epitomised in solving equations of higher degree. InSecondary 2, students are taught to use the method of factorisation to solve simple quadratic equations.Using rational root theorem, we can also attempt to solve equations of higher degree by guessing the roots.

Actually, it is also possible to solve cubic equations just by using technique of factorisation withoutthe use of rational root theorem.

Solve the equation x3 + px+ q = 0.

Recall the factorisation a3 + b3 + c3 − 3abc =1

2(a + b + c)[(a − b)2 + (b − c)2 + (c − a)2]. We shall

use the coefficient determining method to factorise the above equation into this form. The term x3 + qwill represent the a3 + b3 + c3 portion while the term px will represent the term −3abc. We have:

a3 + b3 = q

ab =−p3

We take the cube of the second equation and obtain the following system:

a3 + b3 = q

ab =−p3

27

Now we can solve for a and b since a3 and b3 are now the roots of the quadratic equation z2−qz− p3

27= 0.

We then use the quadratic formula to calculate the value of a3 and b3. We finally derive that:

a =3

√q

2+

√q2

4+p3

27, b =

3

√q

2−√q2

4+p3

27

We just successfully rewrote the original equation in the form of x3 + a3 + b3 − 3xab = 0. Factorising

yields1

2(x + a + b)((x − a)2 + (a − b)2 + (b − x)2) = 0. Hence to obtain x, we just have to solve the

following linear and quadratic equation:

(x+ a+ b) = 0 ((x− a)2 + (a− b)2 + (b− x)2) = 0

Solving yields:

x1 = −a− b, x2,3 =a+ b

2± (a− b)

√3

2i


3.2 Number Theory

Factorisation plays an important role in number theory problems, especially in the discussion of primenumbers and composite numbers. There were many past SMO questions on number theory which madeuse of factorisation. This was a question which appeared in SMO(J) 2nd Round in 2008.

Determine all primes p such that 5p + 4p4 is a perfect square.

To solve this problem, we let 5p + 4p4 = n2 for some positive integer n. We have 5p = (n− 2p2)(n+ 2p2).Since all variables are integers, we infer that n + 2p2 and n − 2p2 are certain powers of 5. So supposen + 2p2 = 5s and n − 2p2 = 5t such that s + t = p and 0 ≤ s < t. We can eliminate n by taking thedifference of the two expressions, hence obtaining 4p2 = 5s(5t−s − 1). Obviously if s > 0, p must be 5and indeed the expression is a perfect square. If s = 0 and t = p, we have 5p = 4p2 + 1. However, we canprove that 5p > 4p2 + 1 if p > 1 using mathematical induction. Hence, the only solution to the problemis p = 5.

3.3 Problem Set

1. Factorise the following expressions:

(a) x4 + x2 + 1

(b) x10 + x5 + 1

(c) x5 + x+ 1

(d) x9 + x4 − x− 1

2. Factorise the following expression:

(a) (a− b)3 + (b− c)3 + (c− a)3

(b) (a+ 2b− 3c)3 + (b+ 2c− 3a)3 + (c+ 2a− 3b)3

(c) (a+ b+ c)3 − a3 − b3 − c3

(d) ab(a2 − b2) + bc(b2 − c2) + ca(c2 − a2)

(e) a2 − 3b2 − 8c2 + 2ab+ 2bc+ 14ca

3. Given that x8 + x7 + x6 + · · ·+ x+ 1 = 0, evaluate x1998 + x1997 + · · ·+ x2 + x+ 1.

4. Find all pairs of integers (x, y) such that x3 + y3 = (x+ y)2.

5. (SMO(J) 2011 First Round P2) It is known that the roots of the equation x5 + 3x4 − 404118x3 −12132362x2 − 12132363x − 20112 = 0 are all integers. How many distinct roots does the equationhave?

6. (SMO(S)2011 First Round P32) It is given that p is a prime number such that x3 +y3−3xy = p−1for some positive integers x and y. Determine the largest possible value of p.

7. (SMO(S)2006 First Round P29) Let a, b be two integers. Suppose x2−x−1 is a factor of polynomialax5 + bx4 + 1. Find the value of a.

8. (SMO(S)2006 First Round P30) How many ordered pairs of integers (x, y) satisfy the equationx√y + y

√x+√

2006xy −√

2006x−√

2006y − 2006 = 0?

9. (SMO(O)2006 First Round P9) Suppose f is a function satisfying f(x + x−1) = x6 + x−6 for allx 6= 0. Determine the value of f(3)

10. (SMO(O)2008 First Round P24) Let f(x) = x3 + 3x+ 1 where x is a real number. Given that theinverse function of f exists and is given by

f−1(x) =

(x− a+

√x2 − bx+ c

2

)1/3

+

(x− a−

√x2 − bx+ c

2

)1/3

where a, b, c are positive constants, find the value of a+ 10b+ 100c.


11. Let r be a real number such that 3√r +

13√r

= 3. Determine the value of r3 +1

r3

12. Prove that for every integer n > 2, the number 22n−2 + 1 is not a prime number.

13. Let a, b, c be positive real numbers satisfying (a2 + b2 + c2)2 > 2(a4 + b4 + c4). Prove that a, b, cmust be the three sides of the same triangle.

14. (IMO 1969) Show that for any positive integers n, there exist infinitely many a such that the numbern4 + a is not prime.


3.4 Solutions

1. (a) x4 + x2 + 1 = x4 + 2x2 + 1− x2

= (x2 + 1)2 − x2

= (x2 − x+ 1)(x2 + x+ 1)

(b) x10 + x5 + 1 =x15 − 1

x5 − 1

=(x3 − 1)(x12 + x9 + x6 + x3 + 1)

(x− 1)(x4 + x3 + x2 + x+ 1)

=(x2 + x+ 1)(x12 + x9 + x6 + x3 + 1)

x4 + x3 + x2 + x+ 1

= (x2 + x+ 1)(x8 − x7 + x5 − x4 + x3 − x+ 1)

(c) x5 + x+ 1 = x5 − x2 + x2 + x+ 1

= x2(x3 − 1) + x2 + x+ 1

= x2(x− 1)(x2 + x+ 1) + (x2 + x+ 1)

= (x3 − x2 + 1)(x2 + x+ 1)

(d) x9 + x4 − x− 1 = x9 + x5 + x4 − x5 − x− 1

= (x4 − 1)(x5 + x+ 1)

= (x− 1)(x+ 1)(x2 + 1)(x3 − x2 + 1)(x2 + x+ 1)

Note: For part (b), (c), (d), we can use complex numbers to show that x2 + x+ 1 is a factor of theexpression. Let us use part (b) to illustrate this. Suppose ω1 and ω2 are the complex roots of theequation x3 − 1 = 0. Since (x− 1)(x2 + x+ 1) = 0, we have ω2

1 + ω1 + 1 = 0 and ω22 + ω2 + 1 = 0.

We note that ω101 + ω5

1 + 1 = ω21 + ω1 + 1 = 0 and ω10

2 + ω52 + 1 = ω2

2 + ω2 + 1 = 0. By factortheorem, we know that x2 + x+ 1 must be a factor of the expression.

2. (a) By using the property that a3 + b3 + c3 = 3abc when a + b + c = 0, we have (a − b)3 + (b −c)3 + (c− a)3 = 3(a− b)(b− c)(c− a).

(b) By using the same property as the previous part, we have (a + 2b − 3c)3 + (b + 2c − 3a)3 +(c+ 2a− 3b)3 = 3(a+ 2b− 3c)(b+ 2c− 3a)(c+ 2a− 3b).

(c) Let a be the principal variable of the function. Suppose a = −b. We have

f(−b) = c3 − b3 + b3 − c3 = 0

By factor theorem, we know that (a + b) is a factor of the expression. Since the expressionis a cyclic expression, (b + c) and (c + a) must be factors of the expressions too. Hence, thefactorised form of the expression must be A(a+ b)(b+ c)(c+ a) where A is a constant. Uponexpansion, we solve that A = 3 and the factorised form of the expression is 3(a+b)(b+c)(c+a).

(d) Let a be the principal variable of the function. Suppse a = b. We have

f(b) = b2(b2 − b2) + bc(b2 − c2)− bc(c2 − b2) = 0

Hence (a− b) is a factor of the expression. Since the expression is a cyclic, (b− c) and (c− a)are also the factors of the expression. Now we suppose a− = b− c. We have

f(−b− c) = (−b− c)[(b+ c)2 − b2] + bc(b2 − c2) + c(−b− c)[c2 − (b+ c)2] = 0

Therefore (a + b + c) is a factor of the expression. The original expression is in the form ofA(a + b + c)(a − b)(b − c)(c − a). Upon expansion and checking the coefficients, we deriveA = −1 and hence the factorised form is (a+ b+ c)(b− a)(c− b)(a− c).

(e) Note that a2 + 2ab− 3b2 = (a+ 3b)(a− 2b). We suppose that the factorised expression of theequation is (a+3b+mc)(a− b+nc) for some constants m,n that we want to determine. Uponexpansion, we obtain a2− 3b2− 8c2 + 2ab+ 2bc+ 14ca = a2− 3b2 +mnc2 + 2ab+ (3n−m)bc+(m+ n)ca. We need to solve the following system of equations:

m+ n = 2

3n−m = 14

mn = −8


Solving yields m = −2, n = 4. Hence the factorised expression is (a+ 3b− 2c)(a− b+ 4c)

3. From the condition we have x9 + x8 + · · ·+ x = 0. Hence, x1998 + x1997 + · · ·+ 1 = (x9 + x8 + · · ·+x)(x1989 + x1980 + · · ·+ x9 + 1) + 1 = 1.

4. Obviously, (k,−k) is a solution for all integers k. Suppose x 6= −y. We have x2 − xy + y2 = x+ y.We can write this expression as x2 − (y + 1)x + y2 − y = 0. By using the quadratic formula, wehave

x =y + 1±

√(y + 1)2 − 4(y2 − y)

2=y + 1±

√6y − 3y2 + 1

2

Hence 6y − 3y2 + 1 ≥ 0 and since y is an integer, y can only be 1 or 2. Hence, the five possiblesolutions are (k,−k), (1, 1), (2, 1), (1, 2), (2, 2)

5. By using the rational root theorem, we know that the roots are either 1,−1, 2011,−2011, 20112 or−20112. Since the sum of roots is −3 while the product of roots is 20112, the only possible set ofsolutions is (−1,−1,−1, 2011,−2011). Hence there are 3 distinct solutions.

6. Introduce the notations σ1 and σ2 to indicate the ESPs for second degree polynomials x + y andxy respectively. We derive:

x3 + y3 − 3xy + 1 = σ31 − 3xy2 − 3x2y − 3xy + 1

= σ31 + 1− 3σ1σ2 − 3σ2

= (σ1 + 1)(σ21 − σ1 + 1)− 3σ2(σ1 + 1)

= (σ1 + 1)(σ21 − σ1 − 3σ2 + 1)

= (x+ y + 1)(x2 + y2 − xy − x− y + 1)

= p

Since x+ y+ 1 > 1, we must have x+ y+ 1 = p and x2 + y2− xy− x− y+ 1 = 1. From the secondequation, we have x2 + y2 − xy − x − y = 0. By using the result in problem 4, we know that theonly solutions are (1, 1), (2, 1), (1, 2), (2, 2). Hence, the largest prime p is 5 when x = y = 2.

7. Suppose the factorised form of the expression is (x2− x− 1)(ax3 + cx2 + dx− 1). Upon expansion,we have ax5 + bx4 + 1 = ax5 + (c−a)x4 + (d− c−a)x3− (1 +d+ c)x2 + (1−d)x+ 1. By comparingcoefficients, we obtain the following system of equations:

d− c− a = 0

1 + d+ c = 0

1− d = 0

Which gives us the solution d = 1, c = −2, a = 3. Hence the solution is a = 3.

8. Let k =√

2006, σ1 =√x+√y and σ2 =

√xy. We have

x√y + y

√x+

√2006xy −

√2006x−

√2006y − 2006 = σ1σ2 + kσ2 − kσ1 − k2

= σ1(σ2 − k) + k(σ2 − k)

= (σ1 + k)(σ2 − k)

= (√x+√y +√

2006)(√xy −

√2006) = 0

Hence we must have xy = 2006. There are 8 possible solutions:

(1, 2006), (2, 1003), (17, 118), (34, 59), (59, 34), (118, 17), (1003, 2), (2006, 1)

9. f(x+ x−1) = x6 + x−6

= (x2 + x−2)(x4 − 2 + x−4)

= [(x+ x−1)2 − 2][(x2 + x−2)2 − 4]

= [(x+ x−1)2 − 2]{[(x+ x−1)2 − 2]2 − 4}Hence, f(x) = (x2 − 2)[(x2 − 2)2 − 4]. We have f(3) = 322.


10. Let y = x3 + 3x+ 1. By shifting all terms to the left we have x3 + 3x+ 1− y = 0. We want to writethis in the form of x3 + a3 + b3 − 3abx = 0 so that we can factorise the expression. By comparingterms, we have:

a3 + b3 = 1− yab = −1

From the second equation, we have a3b3 = −1. Hence, a3 and b3 are the roots of the quadraticequation t2 − (1− y)t− 1 = 0. By using the quadratic formula, we have:

a =3

√1− y +

√y2 − 2y + 5

2b =

3

√1− y −

√y2 − 2y + 5

2

Upon factorisation of x3 + a3 + b3 − 3abx we obtain (x+ a+ b)(x2 + a2 + b2 − ax− bx− ab) = 0.

From the first factor, we obtain x = −a− b =3

√y − 1 +

√y2 − 2y + 5

2+

3

√y − 1−

√y2 − 2y + 5

2.

This suggests that f−1(x) =3

√x− 1 +

√x2 − 2x+ 5

2+

3

√x− 1−

√x2 − 2x+ 5

2, and hence a =

1, b = 2, c = 5. The value of 100c+ 10b+ a is 521.

11. We use the property a3 + b3 + c3 = 3abc when a + b + c = 0 to solve this problem. From the

condition, we have13√r

+ 3√r − 3 = 0. Hence, r +

1

r− 27 = 3( 3

√r)(

13√r

) = −9 and r +1

r= 18.

Similarly, r3 +1

r3− 183 = 3(r3)(

1

r3)(−18) = −54. Therefore r3 +

1

r3= 183 − 54 = 5778.

12. Using the Sophie Germain Identity, we have 1+22n−2 = (1+22

n−2

+22n−1−1)(1−22

n−2

+22n−1−1).

It suffices to prove that 1− 22n−2

+ 22n−1−1 > 1, which is obviously true since 22

n−1−1 > 22n−2

forn > 2.

13. Upon expansion and moving all the terms to the left hand side, we have

(a2 + b2 + c2)2 − 2(a4 + b4 + c4) = 2a2b2 + 2b2c2 + 2c2a2 − a4 − b4 − c4

= −(a2 − b2)2 − c4 + c2(2a2 + 2b2)

= −(a+ b)2(a− b)2 − c4 + c2[(a+ b)2 + (a− b)2]

= −[(a+ b)2 − c2][(a− b)2 − c2]

= (a+ b+ c)(a+ b− c)(a− b+ c)(−a+ b+ c) > 0

Since the expression is larger than 0, we either have all four factors to be larger than 0 or two ofthe factors to be larger than zero. Suppose it is the latter case. Without loss of generality, let usassume that a − b + c < 0 and b + c − a < 0. Taking the sum of these two inequalities, we obtain2c < 0 which is a contradiction. Hence, all four factors of the expression must be larger than zero.We have:

a+ c > b

a+ b > c

b+ c > a

Since the three numbers satisfy the triangle inequality, they can form sides of a triangle.

14. We shall prove that n4 +4k4 is a composite number for all positive integers k. By using the Sophie-Germain Identity, we have n4 + 4k4 = (n2 + 2nk + 2k2)(n2 − 2nk + k2). It suffices to prove thatn2 − 2nk + 2k2 > 1, which is true since (n− k)2 + k2 − 1 > 0. Since there is an infinite amount ofchoices for k, the proposition is proven.

Chapter 4

Miscellaneous Techniques in Algebra

4.1 Discriminant and Vieta’s theorem

The discriminant of a quadratic equation is given by the expression 4 = b2 − 4ac. It is a well-knownfact that a quadratic equation has 2 distinct real roots if 4 > 0, two equal real roots when 4 = 0 andtwo distinct complex roots when 4 < 0. If it is stated that a quadratic equation has two real roots, wecan deduce that 4 ≥ 0. (Note that this condition includes the situation when the two roots are equal. Iremember that I used to regard equal roots as one real root and answered an SMO problem wrongly).

Most students are aware of the application of discriminant to determine the number of real roots. Thediscriminant can also come in handy if it is stated that the roots of the quadratic equation are integersor rational numbers. In both cases, the root of the discriminant must be a rational number i.e. thediscriminant must either be a perfect square or the square of a rational number.

Problem one in the second round of SMO(J)2006 illustrates this idea:

Find all integers (x, y) that satisfy the following equation:

x+ y = x2 − xy + y2

Rearranging yields x2 − (y + 1)x + y2 − y = 0. Since the solutions are integers, this implies that thediscriminant is a perfect square. The discriminant of this equation is 4 = 6y − 3y2 + 1, which is largerthan zero only when y = 0, 1, 2. It turns out that the discriminant is a perfect square for these values,and this gives us the 6 solutions (x, y) = (0, 0), (1, 0), (0, 1), (2, 1), (1, 2), (2, 2).

On the other hand, Vieta’s theorem relates the elementary symmetric polynomials in terms of the rootsof a polynomial to the coefficients of the variable. For a quadratic equation ax2 + bx+ c = 0, we have

x1 + x2 = − ba

x1x2 =c

a

For equations of higher degree such as anxn + an−1x

n−1 + · · ·+ a1x+ a0 = 0, we have:

σ1 =∑sym

xi = x1 + x2 + · · ·+ xn = −an−1

an

σ2 =∑sym

xixj = x1x2 + x1x3 + · · ·+ x2x3 + x2x4 + · · ·+ xn−1xn =an−2

an

...

σn = x1x2 · · ·xn = (−1)na0an

Vieta’s theorem is widely applied in MO problems in various forms. It is also useful to apply Vieta’stheorem if the signs of the roots are given, since the signs of the coefficients are dependent on the signsof each root.

24

CHAPTER 4. MISCELLANEOUS TECHNIQUES IN ALGEBRA 25

4.2 Method of differences

The essence of this method is to break down an expression into differences and eliminate terms by takingtelescoping sum. This method is often applied (but not limited to) if the expression contains fractions orfactorials.

To split a fraction into its corresponding difference expression, we often employ the method of partialfractions. Generally we have:

1.k

(ax+ b)(cx+ d)=

A

ax+ b+

B

cx+ d

2.k

(ax+ b)2=

A

ax+ b+

B

(ax+ b)2

3.k

(ax2 + bx+ c)(dx+ e)=

Ax+B

ax2 + bx+ c+

C

dx+ e

Apart from coefficient determining method, there’s a shortcut that allows us to deduce the constants ofthe numerator of the partial fractions quickly if the denominators are linear expressions. To illustratethis shortcut, consider the problem below.

Find a general formula for the sum1

1× 2× 3+

1

2× 3× 4+ · · ·+ 1

n(n+ 1)(n+ 2)

We want to decompose the fraction 1n(n+1)(n+2) into its partial fractions. To find the constant above

the n term, we can ignore the n part in the original fraction and substitute n = 0 into the fraction. Thisgives us the value 1

2 , which will be the constant term of the partial fraction. Similarly, we ignore theexpression n+ 1 and substitute n = −1 into the fraction, deriving -1 as the constant term of the partialfraction with denominator n+ 1. Finally we ignore the expression n+ 2 and substitute n = −2 into thefraction, obtaining 1

2 as the constant term of the final partial fraction with denominator n+ 2. We have:

1

n(n+ 1)(n+ 2)=

1

2n− 1

n+ 1+

1

2(n+ 1)

By telescoping sum, we have:

1

1× 2× 3+

1

2× 3× 4+ · · ·+ 1

n(n+ 1)(n+ 2)=

1

2− 1

2+

1

4+

1

2(n+ 1)− 1

n+

1

2(n+ 2)

=1

4− 2n2 + 4n+ 1

2n(n+ 1)(n+ 2)

Here’s a problem which involves factorials.

(SMO(S)1999P24) Evaluate 1× 1! + 2× 2! + 3× 3! + · · ·+ 9× 9!.

Note that n× n! = (n+ 1)!− n!. Using this fact, we have

1× 1! + 2× 2! + 3× 3! + · · ·+ 9× 9! = 10!− 1 = 362879.

4.3 Method of fixed ratios

This technique is applicable to questions that has several equal quantities such as a = b = c = d. It isusually more useful for problems with a fixed ratio or quantities in exponential form. Given a conditiona

b=c

d=

e

f, we can let this quantity equate to a constant k and derive a = bk, c = dk, e = fk which

may be useful in solving the problem.

On the other hand, to manage exponential relationships like ax = by = cz, we can equate this quan-tity to another constant k and take logarithm to obtain the relationship x = loga k, y = logb y andz = logc z.


Consider this problem:

(AIME85P7) Suppose a, b, c, d are positive integers that satisfy

a5 = b4

c3 = d2

c− a = 19

Evaluate d− b.

From the first condition, we can let a5 = b4 = e20 for some integer e. Also, we can let c3 = d2 = f6

for some integer f . From the third condition, we have f2 − e4 = 19 ⇔ (f − e2)(f + e2) = 19. Since19 is a prime, we must have f − e2 = 1, f + e2 = 19. Solving yields e = 3 and f = 10. Hence,d− b = 103 − 35 = 757.

4.4 Geometric constructions

This class of problems are very tricky because without the inspiration to use diagrams to simplify theproblem it can be very tedious to solve the problem. One must be very sensitive to algebraic expressionswhich appear like a geometric formula such as the Pythagorean Theorem, cosine rule, equations of circles,etc. The following problem was take from SMO(J) last year:

(SMO(J)2011 First Round P14) Let a, b, c be positive real numbers such that

a2 + ab+ b2 = 25

b2 + bc+ c2 = 49

c2 + ca+ a2 = 64

Find (a+ b+ c)2.

We need to know the value of ab + bc + ca in order to evaluate the quantity (a + b + c)2. It is ob-viously impossible to calculate this value directly from the system of equations. However, we notice thatthe right hand side of the equations are all perfect squares. Also, recall the cosine rule which states thata2 = b2 + c2−2bc cosA for triangle ABC. It looks somewhat similar to the system of equations, providedthat the angle in the cosine rule expression is 120 degrees. Of course, it is impossible to construct atriangle such that each angle is 120 degrees. We can, however, construct three lines from a point suchthat the angle between each line is 120 degrees. Consider the triangle below:

We let |AD| = a, |BD| = b, |CD| = c. Using cosine rule, we derive that |AB| = 5, |BC| = 7, |CA| = 8.We still need to find out the value of ab + bc + ca. Consider the sine formula for area of triangle whichstates that SABC = 1

2bc sinA. We note that the area of the large triangle is equal to the sum of areas ofthe three small triangles. By equating these two quantities using Heron’s formula and sine formula forarea, we have:

1

2(ab+ bc+ ca) sin 120◦ =

√10(5)(2)(3)

ab+ bc+ ca = 40

Finally, we have (a+ b+ c)2 = a2 + b2 + c2 + 2(ab+ bc+ ca) = 12 (25 + 49 + 64 + 3× 40) = 129.


4.5 Problem Set

1. (SMO(S)2009 First Round P3) If two real numbers a and b are randomly chosen from the interval(0, 1), find the probability that the equation x2 −

√ax+ b = 0 has real roots.

2. (SMO(S)2006 First Round P35) Let p be an integer such that both roots of the equation 5x2 −5px+ (66p− 1) = 0 are positive integers. Find the value of p.

3. (AIME2008P7) Let r, s, and t be the three roots of the equation 8x3 + 1001x + 2008 = 0. Find(r + s)3 + (s+ t)3 + (t+ r)3.

4. (AIME2011P15) Let P (x) = x2−3x−9. A real number x is chosen at random from the interval 5 ≤

x ≤ 15. The probability that b√P (x)c =

√P (bxc) is equal to

√a+√b+√c− d

e, where a, b, c, d

and e are positive integers and none of a, b, c, d and e are positive integers. Find a+ b+ c+ d+ e.

5. Compute 1!(12 + 1 + 1) + 2!(22 + 2 + 1) + 3!(32 + 3 + 1) + · · ·+ 9!(92 + 9 + 1).

6. Compute 1× 2 + 2× 3 + 3× 4 + · · ·+ n× (n+ 1).

7. (SMO(J)2006 First Round P9) Find the value of1

3 + 1+

2

32 + 1+

4

34 + 1+ · · ·+ 22006

322006 + 1.

8. (Ukraine) Prove the inequality1√

1 +√

3+

1√5 +√

7+ · · ·+ 1√

9997 +√

9999> 24.

9. (Canada) Givena1b1

=a2b2

=a3b3

and p1, p2, p3 are not all equal to zero. Prove that for all positive

integers n, we have (a1b1

)n =p1a

n1 + p2a

n2 + p3a

n3

p1bn1 + p2bn2 + p3bn3.

10. Solve the following systems of equations:

(a)x2 + y2 + z2

x+ y=

14

3

x2 + y2 + z2

y + z=

14

5

x2 + y2 + z2

z + x=

7

2

(b) ax = by = cz =1

x+

1

y+

1

z(a > 0, b > 0, c > 0)

11. Given that n is a natural number, ax2n+1 = by2

n+1 = cz2n+1, and

1

x+

1

y+

1

z= 1. Prove that

2n+1√ax2n + by2n + cz2n = 2n+1

√a+ 2n+1

√b+ 2n+1

√c.

12. Suppose x1, x2, · · · , x100 are 100 positive numbers that satisfy the following conditions:

(a) x21 + x22 + · · ·+ x2100 > 10000

(b) x1 + x2 + · · ·+ x100 ≤ 300

Prove that it is possible to find 3 numbers such that the sum of these 3 numbers is larger than 100.

13. (AIME2006P15) Given that x, y and z are real numbers that satisfy:

x =

√y2 − 1

16+

√z2 − 1

16

y =

√z2 − 1

25+

√x2 − 1

25

z =

√x2 − 1

36+

√y2 − 1

36

and that x+ y + z =m√n

, where m and n are positive integers and n is not divisible by the square

of any prime, find m+ n.


4.6 Solutions

1. The discriminant of the quadratic equation is a − 4b. Since the equation has real roots, we musthave 4 ≥ 0 and hence a ≥ 4b. The triangular region below indicates the possible values of a and b

such that the inequality holds within the domain a, b ∈ (0, 1). Since the area of the triangle is1

8,

the probability when the equation has real roots is1

8÷ 1 =

1

8

-

XXXXXXXXXXXX0 1

a

6

0

14

1

b

2. Suppose the two roots of the equation are m and n. We have m + n = p. Since m,n are positiveintegers, p must be a positive integer too. Since the quadratic equation has integer coefficients andinteger solutions, the discriminant of the equation must be a perfect square. Suppose 4 = k2 forsome integer k. We have:

(5p)2−4(5)(66p−1) = k2 ⇔ (5p−132)2−k2 = 17404⇔ (5p−132+k)(5p−132−k) = 4×19×229

Since the expressions 5p−132+k and 5p−132−k have the same parity and since 17404 is divisibleby 4, both expressions must be even numbers. We must have:{

5p− 132 + k = 458

5p− 132− k = 38or

{5p− 132 + k = −38

5p− 132− k = −458

The first set of equation gives us p = 76 while the second set of equations give us a negative valueof p. Hence the only solution for p is 76.

3. By using Vieta’s theorem, we have the following relationships:

r + s+ t = 0

rs+ st+ tr =1001

8rst = −251

Since r + s+ t = 0, we have:

(r + s)3 + (s+ t)3 + (t+ r)3 = 3(r + s)(s+ t)(t+ r)

= 3(rs2 + r2s+ st2 + s2t+ tr2 + t2r + 2rst)

= 3[(r + s+ t)(rs+ st+ tr)− rst]= 3(251)

= 753

4. This is a sinister problem that looks complicated but its existence merely serves the purpose ofwasting your time in the competition. Needless to say this brutal problem did serve its purposewell. Firstly, note that P (bxc) must be a perfect square since b

√P (x)c is a positive integer. By

checking all the integers in between 5 and 15 inclusively, we realise that P (x) is a positive integerwhen x = 5, 6, 13.


When x ∈ [5, 6), we have bx2− 3x− 9c = 1. For the equation to hold, the value of x2− 3x− 9 must

be less than 4. By using quadratic equation to solve for x2 − 3x− 9 < 4, we obtain x <3 +√

61

2.

Since this number is smaller than 6, we obtain our first valid interval x ∈ [5,3 +√

61

2).

When x ∈ [6, 7), we have bx2− 3x− 9c = 3. For the equation to hold, the value of x2− 3x− 9 must

be less than 16. By using quadratic equation to solve for x2−3x−9 < 16, we obtain x <3 +√

109

2.

Since this number is smaller than 7, we obtain our second valid interval x ∈ [6,3 +√

109

2).

When x ∈ [13, 14), we have bx2−3x−9c = 11. For the equation to hold, the value of x2−3x−9 must

be less than 144. By using quadratic equation to solve for x2−3x−9 < 144, we obtain x <3 +√

621

2.

Since this number is smaller than 14, we obtain our first valid interval x ∈ [13,3 +√

621

2).

To compute the probability of the equation being valid, we have:

P (b√P (x)c =

√P (bxc)) =

(3 +√

61

2− 5) + (

3 +√

109

2− 6) + (

3 +√

621

2− 13)

10

=

√61 +

√109 +

√621− 39

20

The final solution is 850.

5. Generally, we have

k!(k2 + k + 1) = k![(k + 1)2 − k]

= (k + 1)(k + 1)!− k · k!

By telescoping sum, we have 1!(12 + 1 + 1) + 2!(22 + 2 + 1) + 3!(32 + 3 + 1) + · · ·+ 9!(92 + 9 + 1) =10 · 10!− 1 = 36287999.

6. Generally, we have n(n+ 1) =1

3(n(n+ 1)(n+ 2)− (n− 1)(n)(n+ 1)). By telescoping sum, we have

1× 2 + 2× 3 + · · ·+ n× (n+ 1) =1

3n(n+ 1)(n+ 2)

7. Generally, we have

2k

32k + 1=

2k

32k − 1− 2k

32k − 1+

2k

32k + 1

=2k

32k − 1− 2k+1

32k+1 − 1

By telescoping sum,1

3 + 1+

2

32 + 1+

4

34 + 1+ · · ·+ 22006

322006 + 1=

1

2− 22007

322007 − 1

8. We observe that√

3−√

1

2+

√5−√

3

2+

√7−√

5

2+ · · ·+

√10001−

√9999

2> 48

In addition, we note that1√

1 +√

3+

1√5 +√

7+ · · ·+ 1√

9997 +√

9999>

1√3 +√

5+

1√7 +√

9+

· · ·+ 1√9999 +

√10001

. After rationalising the denominators, we must have

1√1 +√

3+

1√5 +√

7+ · · ·+ 1√

9997 +√

9999> 24

, which proves our proposition.


9. Leta1b1

=a2b2

=a3b3

= k where k is not equal to zero. We obtain three equations a1 = kb1, a2 =

kb2, a3 = kb3. Now we have:

p1an1 + p2a

n2 + p3a

n3

p1bn1 + p2bn2 + p3bn3=p1(kb1)n + p2(kb2)n + p3(kb3)n

p1bn1 + p2bn2 + p3bn3= kn

= (a1b1

)n

which proves the proposition.

10. (a) Let k = x2 + y2 + z2, a = x+ y, b = y + z, c = z + x. From the system of equations, we have14

3a =

14

5b =

14

4c = k, which suggests that a : b : c = 3 : 5 : 4. We let a = 3α, b = 5α, c = 4α

for some constant α. Solving for x, y, z we obtain x = α, y = 2α, z = 3α. Substitute back

into the first equation, we have14α2

3α=

14

3which gives us α = 1. Hence, our solution is

(x, y, z) = (1, 2, 3).

(b) Suppose ax = by = cz = k for some constant k. We have x =k

a, y =

k

b, z =

k

c. We also have

1

x+

1

y+

1

z= k which upon substitution gives us

a+ b+ c

k= k and k = ±

√a+ b+ c. Hence,

our solution is x = ±1

a

√a+ b+ c, y = ±1

b

√a+ b+ c, z = ±1

c

√a+ b+ c.

11. Let ax2n+1 = by2

n+1 = cz2n+1 = k where k is some arbitrary constant. We have ax2n =

k

x, by2n =

k

y, cz2n =

k

zand 2n+1

√ax2n + by2n + cz2n = 2n+1

√k(

1

x+

1

y+

1

z) = 2n+1

√k. On the other hand,

2n+1√a =

2n+1√k

x, 2n+1√b =

2n+1√k

y, 2n+1√c =

2n+1√k

z. The sum of these three expressions, 2n+1

√a +

2n+1√b+ 2n+1

√c = (

1

x+

1

y+

1

z) 2n+1√k = 2n+1

√k. Hence the equality in the question holds.

12. We will prove this problem by contradiction. Let us suppose that it is not possible to find threenumbers such that the sum of these three numbers is more than 100. Without loss of generality,let us assume that x1 ≥ x2 ≥ x3 ≥ · · · ≥ x100. Our assumption suggests that x1 + x2 + x3 < 100.We suppose that x1.x2, x3, · · ·x100 are the sides of 100 squares. Place all these squares in a straightline side by side from x1 to x100. Consider the diagram below

We superpose these squares onto a grid of 3 big squares, each with length 100. Since x1 + x2 +· · · + x100 < 300, the squares will fit within the three squares. Our assumption also suggests thatx1, x2, x3 will fit into the first grid. Now, we divide part of the first grid into three rectangles asshown in the diagram above, each with the length of 100 and height of x1, x2, x3 respectively. Wecan move all the other squares from the second big square into the rectangle with height x2 and


we can move all the squares from the third big square into the rectangle with the height x3. Thissuggests that the sum of the areas of these squares is less than 1002 as they are able to fit withinthe first big square. This contradicts with the condition x21 + x22 + · · ·x2100 > 10000 and hence theassumption is false. This would mean that there must be three numbers (in this case x1, x2, x3)such that their sum is larger than 100.

13. On hindsight, this looks like another AIME problem which serves to bore contestants by forcingthem to square each equations, get rid of radicals and solve the simultaneous quadratic equations.However, there’s an elegant solution which makes use of the following geometric construction.Consider the following triangle with sides AB = x,AC = y,BC = z

From the system of equations and by Pythagorean theorem, it seems that the height of the triangle,

AD,BE,CF are1

6.1

5,

1

4respectively. There are two methods that we can use to solve for x, y, z at

this juncture:

(a) By considering the area of the triangle, we obtain the relationship1

4x =

1

5y =

1

6z. By using

method of fixed ratios, we can let x = 4k, y = 5k, z = 6k for some arbitrary constant k. Uponsubstitution into the original equation, we can solve for k and obtain the value of x, y and z.

(b) Similarly, by using the relationships established by the area of triangle, we have y =5

4x and

z =6

4x. We can use Heron’s formula to set up the following equation:

1

8=

√(15

8x)(

7

8x)(

5

8x)(

3

8x)

which enables us to solve for x easily and derive the value of y and z.

I prefer the second approach. Solving yields x =8

15√

7. Hence x+y+z =

8

15√

7+

10

15√

7+

12

15√

7=

2√7. The final answer is 9.

The above procedure is sufficient for AIME since AIME only requires candidates to give theirfinal answer. However, one requires the following substantiations to prove the uniqueness of thesolution:

(a) To prove that this is the only possible set of altitudes and the only possible triangle that can

be constructed, we observe that the equation x =√y2 − h2 +

√z2 − k2 is strictly increasing

for LHS and strictly decreasing for RHS. Hence, there can only be one possible set of altitudefor triangles with the sides x, y, z based on the system of equations.

(b) To show that there exist a triangle with side lengths x, y, z, we observe that√x2 − k2 <

x,√y2 − k2 < y,

√z2 − k2 < z. From the system of equations, we have x < y + z, y <

z + x, z < x+ y. Hence the three variables satisfy the triangle inequality.

a potpourri of algebra

Documents