My question is ;

Let $n$ be $n=N^{2} m$ , where m is a squarefree integer. Then $n$ can be written that as a sum of two integer squares, if $m$ contains no prime factor of the form $4k+3$.

I have a reference here. Look at this solution please. And notice that m is a integer.

"Suppose that m has no prime factor of the form $4 k+3$ if $m=1,$ then $n=N^{2}+0^{2}$ and we are done. In the case $m>1,le t$ $\mu=p_{1} p_{2} \cdots p_{r}$ be the factorization of m into a product of distinct primes. Each of these primes $p_{i},$ being equal to 2 or of the form $4 k+1,$ can be written as a sum of two squares. Now, the identity

$ \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=(a c+b d)^{2}+(a d-b c)^{2} $ shows that the product of two ( and any finite number by induction) integers, each of which is representable as a sum of two squares, is likewise so representable. Thus, there exist integers $x$ and $y$ satisfying $m=x^{2}+y^{2},$ and so $ \left.n=N^{2} m=N^{2}\left(x^{2}+y^{2}\right)=(N x)^{2}+\ (N y\right)^{2} $ which completes the proof."

Is the solution right for this question?

And What are your different ideas? Thanks.

  • 1
    $\begingroup$ Looks good as long as you assume Fermat's two-square theorem without proof. $\endgroup$ – Clement Yung Jun 26 at 1:32
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    $\begingroup$ Agreed. With a little more work, you can prove it's an if-and-only-if statement. $\endgroup$ – Gerry Myerson Jun 26 at 1:34
  • $\begingroup$ thanks. So what changes can i make to solution and how can i improve it? @Clement yung $\endgroup$ – robert08 Jun 26 at 1:34
  • $\begingroup$ i have its proof (if and only if statement) but the question asks only one statement. so what do you suggest me else? @Gerry Myerson $\endgroup$ – robert08 Jun 26 at 1:37
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    $\begingroup$ I don't think there's any way to prove it without using Fermat, or some equivalent statement about factorization in the Gaussian integers. mathoverflow.net/questions/31113/… may interest you, it refers to people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2323918/… also en.wikipedia.org/wiki/… $\endgroup$ – Gerry Myerson Jun 26 at 1:54

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