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

1. Mar 11, 2006

### neptunecs

Hi
N=a^2+b^2+c^2,
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.
neptunecs

2. Mar 12, 2006

### robert Ihnot

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
3. Mar 13, 2006

### neptunecs

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.

4. Mar 14, 2006

### shmoe

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 $\sum_{n\leq x}r(n)\sim 4/3\pi x^{3/2}$.

5. Mar 16, 2006

### neptunecs

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.
Neptunecs.

6. Apr 22, 2006

### AntonVrba

7. Apr 22, 2006

### mathwonk

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.

8. Apr 25, 2006

### robert Ihnot

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.