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.