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N=a^2+b^2+c^2, a.b.c.N are all integers(>0)

  1. Mar 11, 2006 #1
    a.b.c.N are all integers(>0),I am wondering how many solves for each N.
    Have this problem been solved in Number Theory?
    This problem is originate from a quantum mechanics question I met.
    thanks for any advice.
  2. jcsd
  3. Mar 12, 2006 #2
    Most numbers (integers) can be written as the sum of three squares. However numbers of the form 8k+7 and numbers of the form 4^m(8k+7), m>0, can not be so written. It is easy to show that x^2+y^2+z^2 is not congruent to 7 mod 8.

    All odd squares are congruent to 1 mod 8, and even integers are congruent to 0 or 4. Thus we consider the set of residues {0,1,4}, any three of them can not total 7. http://www.artofproblemsolving.com/Forum/topic-5269.html
    Last edited: Mar 12, 2006
  4. Mar 13, 2006 #3
    Thank you very much.
    It's seems that this problem haven't been solved in Number Theory,it's hard to find a function s(N) to discript the int solutions of N=x^2+y^2+z^2, right?
    Thanks for your opinion and information.
  5. Mar 14, 2006 #4


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    The number of ways to represent an integer N as a sum of three squares has been solved, but it's not simple and involves a series that converges very slowly. It's given in http://www.ma.utexas.edu/~kumchev/P4.pdf for example, and you can find other sources, just look for "sum of three squares". Much nicer asymptotics exist for the summatory function, that is if r(n)=#of ways to express n as a sum of three squares then [itex]\sum_{n\leq x}r(n)\sim 4/3\pi x^{3/2}[/itex].
  6. Mar 16, 2006 #5
    Thank you very much!
    I will read that paper soon,but I think it's difficult for me,because I know only a little about Number Theory ,I major in theoritical physics.......
    But thanks for your answer which make me know that this problem has been solved.
  7. Apr 22, 2006 #6
  8. Apr 22, 2006 #7


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    this is not super hard to prove. it is modeled on gauss(?) proof of fermats thm that primes of form 4k+1 are sums of two squares.

    i.e. if p = 4k+1, then X^2+1 = 0 has a solution mod p. [e.g. mod 5, the solution is 1(2), and mod 13 the solution is 1(2)(3)(4)(5)(6) = 5, mod 13.]

    this means that in (Z/p)[X], the polynomial X^2+1 is reducible, so that the ring (Z/p) is not a domain. but then in Z, the element p must be reducible, so p = (a+bi)(c+di), and taking absolute values of both sides, and squaring, we get p = a^2 + b^2.

    Now Hurwitz defined the integral quaternions in a similar way, Z[i,j,k], (with a judicious denominator of 2), and it follows, similarly that no prime integer p is irreducible in there.

    hence p = (a+bi+cj+dk)(e+fi+gj+hk), and again taking norms of both sides shows, with a little computation, that at least 4p is a sum of 4 squares.

    but euler showed that if n is a sum mof 4 squares so is n/2, and applying this twice, p is a sum of 4 squares.

    in both cases, the product of integers which are sums of 2 or 4 squares, is also. [this is because norms are multiplicative.] so the prime case does it all.

    these proofs rely on the existence of a division algortihm in these two rings, even though the second one is non commutative.
  9. Apr 25, 2006 #8
    mathwonk: so p = (a+bi)(c+di), and taking absolute values of both sides, and squaring, we get p = a^2 + b^2.

    There is a small error here which I point only because it might confuse the beginning student, you mean p = (a+bi)(a-bi)=a^2+b^2.
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