1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: N=a^2+b^2+c^2, a.b.c.N are all integers(>0)
  1. A^2 + b^2 = c^3 (Replies: 8)