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For any value of $y$, is $7|y^2 + 1?$ and $3|y^2 +1$?

as well as

$19|y^2 +1$?

If there is no such $y$, how do you prove it.

Also, I want to know about free software to check these type of divisions in online. Please let me know, if exists such softwares.

user177889
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  • You don't need software to do this...it is basic number theory. We have $p\mid y^2+1$ if and only if $y^2 \equiv -1 \bmod p$. If $p$ is odd then this is equivalent to the Legendre symbol $(-1/p)=1$, which happens if and only if $p\equiv 1 \bmod 4$. – fretty Sep 23 '14 at 11:03
  • Of course if $p=2$ then $y$ is odd. – fretty Sep 23 '14 at 11:04

1 Answers1

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$$y\equiv0,1,2\pmod3\implies y^2\equiv0,1\pmod3\not\equiv-1$$

So, there is no integer $y$ such that $3|(y^2+1)$

See also, $-1$ is a quadratic residue modulo $p$ if and only if $p\equiv 1\pmod{4}$

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lab bhattacharjee
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