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Given an integer $n$, we are asked to investigate about the existence of integer divisors of $n^2+1$ of the form $4k+3$. Can you provide some insights about it?

Martin Sleziak
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curieux_2014
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    If an odd prime number $p$ is a factor of the number $n^2+1$, then it follows that $-1$ is a quadratic residue modulo $p$. This is well known to happen, iff $p\equiv1\pmod4$. The same congruence thus holds for all odd divisors of $n^2+1$. See here for a proof. – Jyrki Lahtonen Jul 09 '16 at 22:05
  • But exactly what is the question. I have problems parsing it. A little bit of testing would quickly show that none of the numbers $2^2+1$, $3^2+1$, $4^2+1$,... have divisors of the form $4k+3$. – Jyrki Lahtonen Jul 09 '16 at 22:07
  • $n^2+1$ has no divisors of the form $4k+3$. – JasonM Jul 09 '16 at 22:11
  • think you for your answers – curieux_2014 Jul 09 '16 at 22:20
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    Related to http://math.stackexchange.com/questions/1843150/prove-that-if-p-equiv-3-pmod4-then-x2y2-not-equiv-0-pmodp. – lhf Jul 09 '16 at 22:27

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If you don't yet know quadratic reciprocity then you can instead use little Fermat. Suppose the prime $\,p = 4k\!+\!3\mid \color{#c00}{n^2\!+1}.\,$ Then $\,p\!-\!1 = 2(2k\!+\!1),\,$ and $\,{\rm mod}\ p\!:\ \color{#c00}{n^2\equiv -1},\,$ so $\,n\not\equiv 0,\,$ so

$${\rm mod}\ p\!:\,\ 1\equiv n^{\large p-1}\equiv (\color{#c00}{n^{\large 2}})^{\large 2k+1}\equiv (\color{#c00}{-1})^{\large 2k+1}\equiv -1$$

Thus $\, 1\equiv -1\,\Rightarrow\,p\mid 2,\,$ contra $\,p = 4k+3\,$ is an odd prime.

Bill Dubuque
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Assume that $n^2+1$ has some divisor $d$ of the form $4k+3$.
Then there is some prime $p\mid d$ that is also of the form $4k+3$, and from $p\mid n$ we have: $$ n^2 + 1 \equiv 0\pmod{p} $$ but that is absurd, since it implies that $-1$ is a quadratic residue $\!\!\pmod{p}$,
while the Legendre symbol $$\left(\frac{-1}{p}\right) = (-1)^{\frac{p-1}{2}} = -1 $$ tells us the opposite.

Jack D'Aurizio
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$n^2+1$ has no algebraic factorization, (meaninig we cannot conclude any factors of $n^2+1$ just yet) the best way to look at this is to find an odd prime $p$ $=$ $3 \pmod 4$, and find a solution to $n^2$ $=$ $-1$ $\pmod p$, which is false since $-1$ is a quadratic nonresidue to any prime $p$ $=$ $3$ $\pmod 4$, on the other hand if $p$ $=$ $1$ $\pmod 4$, then there is a solution to $n^2$ $=$ $-1$ $\pmod p$, and $p$ divides $n^2+1$.

J. Linne
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