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Let $ n,x,y \in \Bbb N$.Show that each divisor of $(n^2+1)$ has got the form $x^2+y^2$.

My first thought: $x^2+y^2|n^2+1$

$\Leftrightarrow n^2\equiv-1\pmod {x^2+y^2}$

here we can use the quadratic residue.

But I thought we should better say:

Let $d \in \Bbb Z$ be a divisor of $(n^2+1)$ we know that if $d=p_1\cdots p_n$ , where $p_i$ is prime, than each $p_i$ is a divisor of $n^2+1$. Than we had to show that each each product of primes can be represented as the sum of two quadratics, right ?

Do can I use the quadratic residue or do I need another important fact ?

I JUST want a hint ! Thanks !

kratos88
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1 Answers1

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  • Due to Lagrange's identity $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$ it is enough to show that any prime divisor of $n^2+1$ is of the form $x^2+y^2$;

  • $2=1^2+1^2$ and if for some odd prime $p$ we have $n^2+1\equiv 0\pmod{p}$, then $-1$ is a quadratic residue $\pmod{p}$ and $p\equiv 1\pmod{4}$;

  • $\mathbb{Z}[i]$ is a UFD / there is a single binary reduced quadratic form of discriminant $-4$. It follows that any prime $p\equiv 1\pmod{4}$ can be represented in a essentially unique way as $x^2+y^2$. For a proof by (Fermat's) descent, have a look at this similar question.

Jack D'Aurizio
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