introducing waring's problem

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1. Sums of squares It’s a classical problem in number theory to ask which integers N can be repre- sented by a sum of squares. After a little bit of playing with sums of two squares, one might be led to conjecture the following theorem. Theorem 1.1 (Fermat). A number N can be represented as a sum of two squares precisely when N is of the form n 2 Q p i where each p i is a prime congruent to 1 mod 4. By looking at squares mod 4, we see that a number is congruent to 3 mod 4 cannot be written as a sum of two squares. This leads us to wonder how many squares do you need to represent every number? By similar congruence considerations(mod 8 now), we discover that you cannot represent any number cogruent to 7 mod 8 as a sum of 3 squares. What about 4 squares? Theorem 1.2 (Lagrange). Every natural number is the sum of at most 4 squares Later, the 3 squares question was solved: Theorem 1.3 (Legendre-Gauss). A number N can be represented as a sum of 3 squares precisely when N is not of the form 4 m (8n - 1) From this we see that the critical number of squares here is 4. Only at this point can we represent every natural number as the sum of 4 squares. This also means that any number can be represented by at most d squares, for any d 4. In particular any number can be represented as the sum of at most 5 squares, at most 6 squares and so on. To see this difference, lets consider the solutions to 5= n 2 1 + n 2 2 + n 2 3 + n 2 4 and 5 = n 2 1 + n 2 2 + n 2 3 + n 2 4 + n 2 5 . To the first question, we have the solutions (0,0,1,2) and all it’s permutations and sign changes. Note that I’m representing a solution as an integer vector in 4 variables. While for the second question, we have (0,0,1,2,0) with all of its permutations and sign changes AND the new solutions (1,1,1,1,1) with all of it’s sign changes and trivial permutations. In particular, we can embed the solutions of 4 squares into the solutions of 5 squares in many natural ways. Because of this we expect the number of possible ways to represent a number as d squares to grow with d, or at least not to decrease. Naturally, this leads to asking how many ways can a number be represented as a sum of squares. By the above embedding, this question doesn’t make sense unless we fix the number of squares. And for 4 squares, we have a nice theorem of Jacobi: Theorem 1.4. Write the number of ways to write a number N N as r(N )= {n Z 4 : n 2 1 + n 2 2 + n 2 3 + n 2 4 = N }. Then r(N )=8 X f |N,46|f f For the purposes of this discussion, I’m not interested in proving Jacobi’s theo- rem. However, I’d like to note that the proof involves showing that the generating series of r(N ) is a modular form of level 4 which allows you to write it as a linear combination of an Eisenstein seris and a cusp form. Computing the first couple of coefficients allows one to do this quickly, then the explicit nature of the Eisenstein series yields the formula. For an exposition see Stein’s ”Complex Analysis”. If I’m not going to prove this nice theorem, then what is the purpose of this discussion? Hopefully by now, I’ve motivated the the question, how many ways can a natural 1

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  • 1. Sums of squares

    Its a classical problem in number theory to ask which integers N can be repre-sented by a sum of squares. After a little bit of playing with sums of two squares,one might be led to conjecture the following theorem.

    Theorem 1.1 (Fermat). A number N can be represented as a sum of two squaresprecisely when N is of the form n2

    pi where each pi is a prime congruent to 1

    mod 4.

    By looking at squares mod 4, we see that a number is congruent to 3 mod 4 cannotbe written as a sum of two squares. This leads us to wonder how many squares doyou need to represent every number? By similar congruence considerations(mod 8now), we discover that you cannot represent any number cogruent to 7 mod 8 as asum of 3 squares. What about 4 squares?

    Theorem 1.2 (Lagrange). Every natural number is the sum of at most 4 squares

    Later, the 3 squares question was solved:

    Theorem 1.3 (Legendre-Gauss). A number N can be represented as a sum of 3squares precisely when N is not of the form 4m(8n 1)

    From this we see that the critical number of squares here is 4. Only at thispoint can we represent every natural number as the sum of 4 squares. This alsomeans that any number can be represented by at most d squares, for any d 4.In particular any number can be represented as the sum of at most 5 squares, atmost 6 squares and so on. To see this difference, lets consider the solutions to5 = n21 + n

    22 + n

    23 + n

    24 and 5 = n

    21 + n

    22 + n

    23 + n

    24 + n

    25. To the first question, we

    have the solutions (0,0,1,2) and all its permutations and sign changes. Note thatIm representing a solution as an integer vector in 4 variables. While for the secondquestion, we have (0,0,1,2,0) with all of its permutations and sign changes AND thenew solutions (1,1,1,1,1) with all of its sign changes and trivial permutations. Inparticular, we can embed the solutions of 4 squares into the solutions of 5 squaresin many natural ways. Because of this we expect the number of possible ways torepresent a number as d squares to grow with d, or at least not to decrease.

    Naturally, this leads to asking how many ways can a number be represented as asum of squares. By the above embedding, this question doesnt make sense unlesswe fix the number of squares. And for 4 squares, we have a nice theorem of Jacobi:

    Theorem 1.4. Write the number of ways to write a number N N as r(N) ={n Z4 : n21 + n22 + n23 + n24 = N}. Then

    r(N) = 8

    f |N,46|ff

    For the purposes of this discussion, Im not interested in proving Jacobis theo-rem. However, Id like to note that the proof involves showing that the generatingseries of r(N) is a modular form of level 4 which allows you to write it as a linearcombination of an Eisenstein seris and a cusp form. Computing the first couple ofcoefficients allows one to do this quickly, then the explicit nature of the Eisensteinseries yields the formula. For an exposition see Steins Complex Analysis. If Imnot going to prove this nice theorem, then what is the purpose of this discussion?Hopefully by now, Ive motivated the the question, how many ways can a natural

    1

  • 2number be written as the sum of d squares for d 4. In subsequent posts, I willintroduce the HardyLittlewood circle method to study asymptotics for fixed d 5and the Kloosterman refinement to reduce to 4 squares. From there well look atrelated equidistribution and ergodic theorems. Ill prove

    Theorem 1.5 (HardyLittlewood). Let rd(N) = {x = (x1, . . . , xd) Zd :d

    i=1 x2i =

    N}. For d 5,rd(N) = S(N)Nd/21 + o(Nd/21)

    where S(N) is a constant depending on N lying in an interval [a, b] (0,)Then extend this to

    Theorem 1.6 (HardyLittlewood). Let r4(N) = {x = (x1, . . . , x4) Z4 :4

    i=1 x2i =

    N}.rd(N) = S(N)N + o(N)

    where S(N) is a constant depending on N . (Note that d = 4 so d/2 1 = 1.)Theorem 1.7. For a point x = (x1, . . . , xd) Rd on a sphere of radius

    N

    e.g.d

    i=1 x2i = N , we can project this point onto the unit sphere by rescaling,

    x 7 x/N . In dimensions d 5, for every natural number N , we have SN , thesphere of radius

    N . Consider the lattice points on SN projected onto the unit

    sphere. This sequence of points become equidistributed on the unit sphere.

    2. Warings problem

    Now that weve discussed sums of squares for a bit, you might be wonderingwhat happens if you changed squares to cubes or fourth power, etc. In 1770, thesame year as Lagranges 4 square theorem, Waring asserted that that for any degreek 2, there is some dimension g(k) such that every number is the sum of g(k) kthpowers. Hilbert was the first to prove this, but his bounds were very weak andgave no idea what the minimum dimension possible is. HardyLittlewood lookedat Warings problem and using their circle method were able to give reasonablebounds for the minimal dimension, g(k). A feature of Warings problem is smallnumbers need a lot of summands. For instance, 2k 1 must be written as thesum of 2k 1 1s. And like for squares there can be congruence obstructions thatforce g(k) to be at least a certain size. This led HardyLittlewood to studying avariant of Warings problem where one only needs all sufficiently large numbers tobe represented as a sum of kth powers; the minimal dimension here is denoted G(k).For this problem, they gave an excellent result:

    Theorem 2.1 (HardyLittlewood). For k 2, let rk,d(N) = {x = (x1, . . . , xd) Zd :

    di=1 x

    ki = N}. If d k2k1, then

    rd(N) = S(N)Nd/k1 + o(Nd/k1)

    where S(N) is a constant depending on N lying in an interval [a, b] (0,)Hua made a substantial improvement by reducing g(k) to 2k while Vinogradov

    made the most impressive improvement showing that g(k) is O(k3) as k tends toinfinity.

    Remarks 1. By dimension, I mean the number of summands. 2. The notationg(k) and G(k) were introduced by Hardy-Littlewood.

  • 3Ill prove Hardy-Littlewoods result which will be easy after the sums of squarescase and then Ill make use of Huas inequality to get Huas result. Unfortunately,I wont have time to cover Vinogradovs results especially since I dont understandthem.

    Im not sure how this will turn out and so my goals may change; I will try toreflect any changes on this page.

    Left to do: 1. Add references and links 2. Correct typos 3. Do I want a list oflectures on this page?