math inves-pythagorean theorem

8/2/2019 Math Inves-pythagorean Theorem

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Supplements and Links

Pythagorean triple finder (TI-86): Finds all primitive Pythagorean triples withhypotenuses within a given range.

Pythagorean triple generator (TI-86): Gives the Pythagorean triple generated by twonumbers.

List of Primitive Pythagorean Triples Pythagorean Triples Project

I. Introduction

The Pythagorean theorem carries the name of the Greek mathematician Pythagoras, who lived inthe 6th century BCE, though the theorem had been known elsewhere for some time before.

Although the Pythagorean theorem arose in geometry, we will be concerned strictly with thenumber theoretic properties of the Pythagorean equation, using the connection to geometry only

as a jumping off point.

Theorem 1(Pythagorean Theorem and converse) Let x, y, and z be positive reals. Then z is thelength of the hypotenuse of a right triangle with side lengths x, y, and z if and only if

(1)

A Pythagorean triple is an ordered triple (x,y,z) of three positive integers such thatx2

+y2

=z2.

Ifx,y, andz are relatively prime, then the triple is calledprimitive.

Let us first note the parity ofx,y, andz in primitive triples, that is their valuesmodulo2. Since

02 0, 12 1, 22 0, and 32 1 mod 4, the only squares modulo 4 are 0 and 1. LettingX=x2, Y

=y2, andZ=z2, we have the following solutions toX+ YZmod 4:

0 + 0 00 + 1 11 + 0 1

(1 + 1 2 is not a solution because 2 is not a square modulo 4.) The first of these solutionscorresponds only to nonprimitive triplesx,y, andz are all divisible by 2. Therefore anyprimitive triple corresponds to one of the other two solutions. In either case,Zis odd, and exactlyone ofXand Yis always odd. Thusz is odd, and exactly one ofx andy is odd. To establish a

convention, let us say thatx is always odd andy is always even, since otherwise we can rename

the variables of a given triple to obtain this. A primitive Pythagorean triple therefore has a
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unique representation (x,y,z), wherey is even andx andz are odd.

Let a =zx and b =zy. Substituting forx andy into(1)gives

or In fact, only one of these values forz holds. The solution

results inx ory being negative, so we take

(2)

Define

Then the following hold:

(3)

Solving for rand for s, we find

(4)

The generators rand s will provide the basis of study for primitive triples. Most texts use the

more common generators m and n, wherex = m2n2,y = 2mn, andz = m2 + n2. Though I believe

rand s as defined above are more natural choices (since they emphasize symmetry in thegenerators and not in the side lengths), the analysis is the same in the end, as m = r+ s and n = r.

II. Properties ofr, s, x, y, and z

Theorem 2A Pythagorean triple is primitive if and only if r and s are integers, s is odd, and(r,

s) = 1.

Proof. Let (x,y,z) be a primitive Pythagorean triple (with even y). The quantitiesz +y and 2y

are relatively prime becausez andy are relatively prime in primitive triples andz +y is odd. We
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have

soz +y andzy are relatively prime (because any common divisor would also divide 2y). ThePythagorean relation(1)implies that

soz +y andzy must both be squares because they are relatively prime. Thereforeis an odd integer. By(4),ris also an integer because

is even.Now that we have established that rand s are integers for primitive triples, we will show thatthey are relatively prime. Let m = (r, s). Since m divides both rand s, m dividesx,y, andz by(3).

But sincex andy (in particular) are relatively prime, this means m = 1.

For the converse, assume s is odd and (r, s) = 1. Because , we have that (r,z)= 1, since any divisor of both randz is a divisor ofs but rand s are relatively prime. By the

definition ofr,x + 2r2

=z, so similarly (x,z) = 1. Therefore the triple (x,y,z) is primitive.

CorollaryThe number of primitive Pythagorean triples is countably infinite.

Proof. Since each relatively prime pair (r, s) with odd s gives a unique primitive triple, it sufficesto consider the cardinality of the set S = {(r, s) | rand s are relatively prime and s is odd}. This

set is infinite since for each odd s there is at least one rrelatively prime to s, but S is no larger

than the rationals (by mapping S injectively into the rationals by sending (r, s) to r/s). Thereforethe set Sis countably infinite.

Next we will prove several results regarding triples modulo 3, 4, and 5.

Remark 1If(x,y,z) is a primitive Pythagorean triple, then y 0 mod 4,z 1 mod 4, exactlyone of{x,y} satisfies k 0 mod 3, and exactly one of{x,y,z} satisfies k 0 mod 5.

Proof. From(3),y = 2r(r+ s). Either ris even or r+ s is even (since s is odd), so 4 dividesy.As shown above, the only squares modulo 4 are 0 and 1. Sincez = r2 + (r+ s)2 is a sum of two

squares and s is odd, there are two possibilities. Ifris odd thenz 1 + 0 1 mod 4, and ifriseven thenz 0 + 1 1 mod 4.

The only squares modulo 3 are 0 and 1. WithX=x2, Y=y

2, andZ=z

2, the congruenceX+ YZ
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mod 3 has the following solutions:

0 + 0 00 + 1 1

1 + 0 1

The first case is nonprimitive, so eitherx2

ory2

is divisible by 3, and this implies that eitherx oryis divisible by 3.

The only squares modulo 5 are 0, 1, and 4, soX+ YZmod 5 has solutions (up to permutationofXand Y):

0 + 0 00 + 1 1

0 + 4 41 + 4 0

Again, the first case is nonprimitive so exactly one side is divisible by 5.

To convince yourself that these are the only three generalizations we can make like this,

experiment with different prime moduli with theTripes ModulopJavaScript program. Larger

numbers (7, 11, 13,...) yield no obvious statements about the divisors of the respective triplemembers.

III. Counting Triples

DefinitionTwo triples are siblings if they have a common hypotenuse.

The first such pair of primitive triples stem from the hypotenuse 65 (332 + 562 = 632 + 162 =

652):

x y z r s

33 56 65 4 3

63 16 65 1 7

There are also primitive siblings of the hypotenuse 85:

x y z r s
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13 84 85 6 1

77 36 85 2 7

If we search larger numbers, we can find larger sets of siblings. There are, for example, four

primitive triples with hypotenuse 1105:

x y z r s

47 1104 1105 23 1

817 744 1105 12 19

943 576 1105 9 23

1073 264 1105 4 29

In general, the number of primitive Pythagorean triples of hypotenuse n is dependent on thenumber of prime factors ofn that are congruent to 1 modulo 4. It turns out that only powers of 2

appear as these numbers. The numbers in thesequence5, 65, 1105, 32045, 1185665, 48612265,2576450045,... are the smallest hypotenuses that harbor 1, 2, 4, 8, 16, 32, 64,... primitive

Pythagorean triples. The terms are given by multiplying successive primes that satisfyp 1 mod4:

The hypotenuse 32045 has eight primitive triples:

x y z r s

2277 31964 32045 122 9

8283 30956 32045 109 33

17253 27004 32045 86 7121093 24124 32045 74 89

23067 22244 32045 67 99

27813 15916 32045 46 127

31323 6764 32045 19 159
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32037 716 32045 2 177

But why do primitive triples form only in sets of 2, and what is the connection to primesp 1

mod 4? Let us look at the prime factors ofn more closely.

By examining thedata, it appears that the prime factors of each hypotenuse satisfyp 1 mod 4.Further, each prime hyptenuse holds only one triple, and the more factors a hypotenuse has, the

more triples it holds. Specifically, if(n) is the number of distinct prime factors ofn and all of

the prime factors ofn satisfyp 1 mod 4, then n appears to be the hypotenuse of exactly 2(n)1primitive Pythagorean triples. (Ifn has any prime factorsp 2 or 3 mod 4, then n is not thehypotenuse of any primitive triples.) In light of this, it makes sense that the first sibling sets of

order 2

will be found on the multiples of the first primes that satisfyp 1 mod 4.

Of course, explaining where 2(z)1 arises is more tricky. Some experimentation shows that, for

n with the above criteria,(3)has 2(n)1 positive integral solutions for rand s. We are interested

in solutions to n = r

2

+ (r+ s)

2

where rand r+ s are relatively prime (or, equivalently, (r, s) = 1).To see what is happening, first let us quote a theorem attributed to Fermat, the proof of which we

leave to a book covering introductory number theory.

Theorem 3(Fermat) Let p be a prime such that p 1 mod 4. Then p can be expressed uniquelyas a sum of two squares.

By "uniquely" it is meant that there exist unique integers a and b such that 0 < a < b andp = a2 +b

2. Using Fermat's theorem, we decompose a number n into is prime factors and write each

prime as a sum of two squares. We then make use of theFibonacci identity,

(5)

by which we can write the product of sums of two squares again as the sum of two squares.Multiplying together the sum of squares representations for the prime factors ofn gives a sum of

squares representation for n, and since the hypotenuse of a primitive triple is a sum r2

+ (r+ s)2

of two squares (by(3)), we will be able to generate primitive triples of hypotenuse n simply by

knowing the sum of squares representations of the prime factors ofn.

For example, for n = 1105,

This solution corresponds to r= 23, s = 1, which generate the triple (47, 1104, 1105). The othersolutions to 1105 = r2 + (r+ s)2 can be obtained by changing the order in which the factors are

multiplied together and by reversing the sum of squares expressions of some factors.
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Actually counting the solutions that we get in this way is cumbersome, however. The system is

not a very natural one, and "Fibonacci multiplication" is in fact not even associative. Much betterare the Gaussian integerscomplex numbers of the form a + bi, where a and b are ordinaryintegers and i

2=1. By Fermat's theorem, each prime factorp 1 mod 4 ofn conveniently has

the representationp = a2

+ b2

= (a + bi) (abi). Multiplying two Gaussian integers exactlyresults in the Fibonacci identity:

Rather than multiplying sum of squares representations, we will multiply Gaussian integers. We

obtain the four solutions to 1105 = r2

+ (r+ s)2

by multiplying (Gaussian) "halves" of its(integer) factorsp = (a + bi) (abi) . For each of the three primes dividing 1105, we have twochoices of which Gaussian factor to take. Complex conjugation of a solution does not change the

sum of squares representation, so we can choose the first such factor to have positive imaginary

part. The solutions are then

By multiplying each by its complex conjugate, these correspond to the four representations of1105 as a sum of two relatively prime squares (and thus to the four primitive triples with

hypotenuse 1105).

PropositionLet n > 1 be a natural number whose prime factors all satisfy p 1 mod 4. Then thenumber of distinct ways that n can be expressed as a sum a2 + b2 of two squares with a and b arerelatively prime is 2

(n)1, where (n) is the number of distinct prime factors of n.

This result allows us to count triples of a given hypotenuse: Because z has 2(z)1 sum of squares

representations when all its factors satisfyp 1 mod 4, there are 2(z)1 distinct solutions to(3).In each of these solutions, rand s are relatively prime (because r+ si is not divisible by anyinteger prime) and s is odd (because an even s results in an evenz). Thereforez is the hypotenuse

of exactly 2(z)1 primitive Pythagorean triples. Not only does it allow counting, however, but it

provides an algorithm to actually construct the triples given the sums of square representations of

its prime factors. A simple program can readily compute, for example, the 214

= 16384 primitive

triples with hypotenuse n = 16880412096169215173498945, which is the product of the firstfifteen primes that satisfyp 1 mod 4.

Until now we have considered triples based only onz; with what we have learned it is now easy

to understand them sorted also byx or byy (where "P" denotes a prime leg and "P" denotes aprime power leg):
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x y z r s x y z r s x y z r s x y z r s

3 P 4 51

1

2

120 29 2

3

3 4

P

51

1 45

2

853

2

5

5 P 12 132

1

2

1

22

0

22

1

1

0

1

15 8P

171

319

5

2

8

19

7

1

1

3

7 P 24 2531

23

P264

265

11

1

512

132

1255

32

P

257

1

15

9P

40 41

4

1

2

5P

31

2

31

3

1

2

1

35

1

237

1

5 77

3

685

2

7

1

1P 60 61

5

1

2

7P

36

4

36

5

1

3

1

63

1

6P

65

1

7

32

3

3

6

32

5

1

1

7

13

P 84 8561

29

P420

421

14

1

2120

292

3 940

414

1

1

58 17

1

3

3

1P

48

0

48

1

1

5

1

99

2

0

10

1

1

9

39

9

4

0

40

1

1

1

9

15

112

113

71

33

56 65 43

724

253

1117

44

125

2

9

1

7P

14

4

14

5

8

1

3

3

54

4

54

5

1

6

1

14

3

2

4

14

5

1

1

1

48

3

4

4

48

5

1

2

1

1

9P

18

0

18

1

9

1

Very similar to the case ofz, the prime factors of a givenx determine how many triples havex asa leg (without the restriction that the prime factors must satisfyp 1 mod 4): Ifx 1 mod 2,thenx is a leg of 2

(x)1 primitive triples. And ify 0 mod 4, theny is a leg of 2(y)1 primitivetriples.

Theorem 3Let (n) be the number of distinct prime factors of a natural number n > 1. The

number of primitive Pythagorean triples of which n is a member is

(8)

We now generalize the problem of counting to include nonprimitive triples. (For a nonprimitive

triple, rand s are irrational unless the common factor of the triple is a square.)


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Theorem 4If the natural number z decomposes into distinct primes as

with pi 1 mod 4 for all i and qj 3 mod 4 for all j, then z is the hypotenuse of exactly

(9)

Pythagorean triples.

Proof. Letz be a natural number. Since primes q 3 mod 4 and the prime 2 do not contribute toprimitive triples, these factors inz do not contribute to the total number of triples with

hypotenusez (i.e., removing all of these factors from a given z results in an integer that is thehypotenuse of the same number of triples asz). Therefore it suffices to consider natural numbers

of the form withpi 1 mod 4 for each i. The number of divisors ofz thatcan be written as a product of powers of exactlyj primes is j(1,2,...,k), thejthelementary

symmetric polynomialon the exponents 1, 2, ..., k. Each such divisor dofz is the hypotenuseof 2

(d)1 = 2j1 primitive triples, byTheorem 3, and so contributes 2j

1 nonprimitive triples of

hypotenusez by simply multiplying byz /d. Thus, by a property of the elementary symmetric

polynomials,

For example, 3142875 = 3 53

172

29 is the hypotenuse of exactly T(3142875) = ((2 3 + 1)(2 2 + 1) (2 1 + 1)1) / 2 = 52 Pythagorean triples.

CorollaryLet z = Q p1p2 ...pk> 1 be a natural number with exactly k distinct prime factors

satisfying p 1 mod 4 and no factor of the form p2 for any prime p 1 mod 4. Then the numberof Pythagorean triples with hypotenuse z is T(z) = (3k1) / 2.
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IV. Restrictions on Primitive Triples

In this section we will be interested in triples that satisfy certain conditions. Sometimes a triplethat has a pair of sides with only a unit difference is referred to as a "twin." For example, the

triples (3, 4, 5), (5, 12, 13), (7, 24, 25), and (21, 20, 29) each have a pair of sides that differ by 1.By browsingdata, one finds that triples satisfyingzy = 1 are more common than thosesatisfying |xy| = 1. There are infinitely many of both types, but the latter much less denselypopulate the world of primitive triples.

To find triples with b =zy = 1, we solvex2 + (z1)2 =z2 forz to find

(10)

Any oddx then gives such a triple. Substituting this into the definition ofrgives

Since s= b = 1, the first few triples with b = 1 are:

x y z r s

3 4 5 1 1

5 12 13 2 17 24 25 3 1

9 40 41 4 1

11 60 61 5 1

13 84 85 6 1

Moreover, the sequence ofy values is 4 times thetriangular numbers, since, by(3),

This same procedure can be used for any valid a or b value because the resulting equation willalways be linear inz. From(3)we see that the nth triples with fixed a and b values are given by
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(11)

We now turn to primitive triples with the restriction that |xy| is a given difference d. Thefollowing is a list of the first few triples for which d= 1. (The first twenty-five of these triples

arelisted.)

x y z r s xy

3 4 5 1 1 1

21 20 29 2 3 1119 120 169 5 7 1

697 696 985 12 17 1

4059 4060 5741 29 41 1

To find the general form of these triples we must resort to a new technique, for the equation x2 +

(x + d)2

=z2

does not lead to a natural assignment of an index to one of the generators. But sincerand s display simple patterns in fixed-a and -b triples, we might suspect that they will for fixed-dtriples also. We notice two (among many) recurrence relations that seem to hold: rn = rn1 + sn

1 and sn = rn + rn1. Substituting for sn1 we find

Second-order recurrence relations have solutions of the form rn = kn. Thus, from the recurrence

relation,

the nonzero solutions of which satisfy k22k1 = 0. So we have

Both of these values for ksatisfy the recurrence, but we want the first and second terms to be 1and 2 (the rvalues of the first two triples). Since a linear combination of the two values for kalso

satisfies the recurrence, we construct the general solution for rn,
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and solve for the constants we introduce by specifying the first two terms:

We find that

Then

and by the recurrence relations,

So the nth triple satisfying |xy| = 1 is given (from(3)) by

(12)

Clearly the difference ofxn andyn is always 1. Intuitively, we know that as n gets large, the

triples (xn,yn,zn) approach the side lengths of a right isosceles triangle, the ratios zn /xn andzn /

ynapproaching 2. It is interesting, however, to examine the ratiosn/rnas n gets large; one findsthat not only does this ratio also approach 2, but sn /rn is the nth continued fraction convergentto 2. This connection becomes apparent when we approach the solution of |xy| = 1 triplesanother way and obtain aPell equation, the solutions of which are these convergents:
http://www.math.rutgers.edu/~erowland/pythagoreantriples.html#equation3http://www.math.rutgers.edu/~erowland/pythagoreantriples.html#equation3http://www.math.rutgers.edu/~erowland/pythagoreantriples.html#equation3http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=001333http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=001333http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=001333http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=001333http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=000129http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=000129http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=000129http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=000129http://mathworld.wolfram.com/PellEquation.htmlhttp://mathworld.wolfram.com/PellEquation.htmlhttp://mathworld.wolfram.com/PellEquation.htmlhttp://mathworld.wolfram.com/PellEquation.htmlhttp://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=000129http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=001333http://www.math.rutgers.edu/~erowland/pythagoreantriples.html#equation3


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or

Because the sequence sn /rnare the continued fraction convergents to 2, rn and sn are relativelyprime. Moreover, s is odd, and so rand s generate a primitive triple.

When d 1, the situation is slightly more complicated. Take d= 7 for example. The first fewtriples are:

x y z r s xy

5 12 13 2 1 7

15 8 17 1 3 755 48 73 3 5 7

65 72 97 4 5 7

297 304 425 8 11 7

403 396 565 9 13 7

1755 1748 2477 19 27 7

2325 2332 3293 22 31 7

10205 10212 14437 46 65 7

13575 13568 19193 53 75 7

At first glance it appears that these rand s values hardly form nice recurrence relations. But in

fact they do; there are two "threads" of the same relation (with different initial values) running

through these triples, more easily seen when they are sorted:

x y z r s xy x y z r s xy

5 12 13 2 1 7 15 8 17 1 3 7

55 48 73 3 5 7 65 72 97 4 5 7

297 304 425 8 11 7 403 396 565 9 13 71755 1748 2477 19 27 7 2325 2332 3293 22 31 7

10205 10212 14437 46 65 7 13575 13568 19193 53 75 7

Again the recurrence is


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Therefore, as above, the general solution is

(13)

So the first thread of triples for which |xy| = 7 is given by

and the second thread is given by

In general,xy = s2

2r2

, and the only numbers of this form, for rand s relatively prime and sodd, are those whose prime factors all satisfyp 1 mod 8. These primes comprise the sequence7, 17, 23, 31, 41, 47, 71, 73, 79, 89, 97,... As representations as a sum of two squares werecounted by factoring over Z[i] inSection III, we can count "fundamental" representations of a

given number as s22r2 = (s + r2) (sr2) by factoring overZ[2]. For example, 1 = 122

12. The ring Z[2] has infinitely many units, given by (1 + 2)n for integers n. Call a

representation s22r2fundamental ifs ror s 2r. Each fundamental representation s122r1

2

gives rise to a thread of triples generated by sn + rn2 = (s1 + r12) (1 + 2)n for nonnegative
http://www.math.rutgers.edu/~erowland/pythagoreantriples.html#sectionIIIhttp://www.math.rutgers.edu/~erowland/pythagoreantriples.html#sectionIIIhttp://www.math.rutgers.edu/~erowland/pythagoreantriples.html#sectionIIIhttp://www.math.rutgers.edu/~erowland/pythagoreantriples.html#sectionIII


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integers n. Each prime factor contributes two fundamental solutions, so we have the following

result.

Proposition 2Let d be a natural number whose prime factors all satisfy p 1 mod 8. Then thenumber of fundamental representations of d in the form s

22r

2for positive integers r and s is

2(d), where (d) is the number of distinct prime factors of d.

(Alternatively, we could have considered sequences of triples that are infinitely long in both the

positive and negative directions. In this case, there are 2(d)1 (double-sided) threads for a given

d, and the case d= 1 must be separated from the others (since it still has only one thread, which

is redundant). In this setup, however, we obtain triples with negative legs.)

Each fundamental solution (r1, s1) gives rise to the sequence of triples

For example, let d= 2737 = 7 17 23. The fundamental representations for these prime factorsare

Thus 2737 has the following fundamental representations (where we divide each product by a

power of 1 + 2 to obtain fundamental representations, in this case the first power):


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An additional four threads come from the fundamental representations of2737:

References

Relevant links:

Pythagorean Tripleon MathWorld Right Triangleon MathWorld NOVA: The Proof:NOVA's documentary about Andrew Wiles's proof of Fermat's Last

Theorem

Power Page:accurately subtitled "Interesting Stuff About Powers of Numbers" Pythagorean Triplets:all about triples that satisfyzy = 1

Project Expositions

Pascal's SimplicesPythagorean TriplesRegular Polygons

Regular PolyhedraRegular PolytopesSums of Consecutive Powers

Mathematical InductionModular ArithmeticPolynomial Equations

InvestigationsCalculatorsPopular Books

Eric S. Rowland
http://mathworld.wolfram.com/PythagoreanTriple.htmlhttp://mathworld.wolfram.com/PythagoreanTriple.htmlhttp://mathworld.wolfram.com/RightTriangle.htmlhttp://mathworld.wolfram.com/RightTriangle.htmlhttp://www.pbs.org/wgbh/nova/proofhttp://www.pbs.org/wgbh/nova/proofhttp://www.uwgb.edu/dutchs/recmath/rmpowers.htmhttp://www.uwgb.edu/dutchs/recmath/rmpowers.htmhttp://www.friesian.com/pythag.htmhttp://www.friesian.com/pythag.htmhttp://www.math.rutgers.edu/~erowland/pythagoreantriples-project.html#footerhttp://www.math.rutgers.edu/~erowland/investigations.html#expositionshttp://www.math.rutgers.edu/~erowland/investigations.html#expositionshttp://www.math.rutgers.edu/~erowland/investigations.html#expositionshttp://www.math.rutgers.edu/~erowland/pascalssimplices.htmlhttp://www.math.rutgers.edu/~erowland/pascalssimplices.htmlhttp://www.math.rutgers.edu/~erowland/pythagoreantriples.htmlhttp://www.math.rutgers.edu/~erowland/pythagoreantriples.htmlhttp://www.math.rutgers.edu/~erowland/pythagoreantriples.htmlhttp://www.math.rutgers.edu/~erowland/polygons.htmlhttp://www.math.rutgers.edu/~erowland/polygons.htmlhttp://www.math.rutgers.edu/~erowland/polygons.htmlhttp://www.math.rutgers.edu/~erowland/polyhedra.htmlhttp://www.math.rutgers.edu/~erowland/polyhedra.htmlhttp://www.math.rutgers.edu/~erowland/polytopes.htmlhttp://www.math.rutgers.edu/~erowland/polytopes.htmlhttp://www.math.rutgers.edu/~erowland/polytopes.htmlhttp://www.math.rutgers.edu/~erowland/sumsofpowers.htmlhttp://www.math.rutgers.edu/~erowland/sumsofpowers.htmlhttp://www.math.rutgers.edu/~erowland/sumsofpowers.htmlhttp://www.math.rutgers.edu/~erowland/induction.htmlhttp://www.math.rutgers.edu/~erowland/induction.htmlhttp://www.math.rutgers.edu/~erowland/modulararithmetic.htmlhttp://www.math.rutgers.edu/~erowland/modulararithmetic.htmlhttp://www.math.rutgers.edu/~erowland/modulararithmetic.htmlhttp://www.math.rutgers.edu/~erowland/polynomialequations.htmlhttp://www.math.rutgers.edu/~erowland/polynomialequations.htmlhttp://www.math.rutgers.edu/~erowland/polynomialequations.htmlhttp://www.math.rutgers.edu/~erowland/investigations.htmlhttp://www.math.rutgers.edu/~erowland/investigations.htmlhttp://www.math.rutgers.edu/~erowland/calculators.htmlhttp://www.math.rutgers.edu/~erowland/calculators.htmlhttp://www.math.rutgers.edu/~erowland/calculators.htmlhttp://www.math.rutgers.edu/~erowland/books.htmlhttp://www.math.rutgers.edu/~erowland/books.htmlhttp://www.math.rutgers.edu/~erowland/books.htmlhttp://www.math.rutgers.edu/~erowland/index.htmlhttp://www.math.rutgers.edu/~erowland/index.htmlhttp://www.math.rutgers.edu/~erowland/index.htmlhttp://www.math.rutgers.edu/~erowland/books.htmlhttp://www.math.rutgers.edu/~erowland/calculators.htmlhttp://www.math.rutgers.edu/~erowland/investigations.htmlhttp://www.math.rutgers.edu/~erowland/polynomialequations.htmlhttp://www.math.rutgers.edu/~erowland/modulararithmetic.htmlhttp://www.math.rutgers.edu/~erowland/induction.htmlhttp://www.math.rutgers.edu/~erowland/sumsofpowers.htmlhttp://www.math.rutgers.edu/~erowland/polytopes.htmlhttp://www.math.rutgers.edu/~erowland/polyhedra.htmlhttp://www.math.rutgers.edu/~erowland/polygons.htmlhttp://www.math.rutgers.edu/~erowland/pythagoreantriples.htmlhttp://www.math.rutgers.edu/~erowland/pascalssimplices.htmlhttp://www.math.rutgers.edu/~erowland/investigations.html#expositionshttp://www.math.rutgers.edu/~erowland/pythagoreantriples-project.html#footerhttp://www.friesian.com/pythag.htmhttp://www.uwgb.edu/dutchs/recmath/rmpowers.htmhttp://www.pbs.org/wgbh/nova/proofhttp://mathworld.wolfram.com/RightTriangle.htmlhttp://mathworld.wolfram.com/PythagoreanTriple.html


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math inves-pythagorean theorem

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