Using Wilson's theorem, we show that if $p$ is an odd prime $x^2 + 1 \equiv 0 \pmod p$ has a solution if and only if $p\equiv 1\pmod 4$ . Why cant $p$ be of the form $4n+3$ ?
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$-1$ is not a quadratic residue modulo $p$ , if $p$ is of the form $4k+3$. Hence such a prime cannot divide a number of the form $x^2+1$ with integer $x$. – Peter Sep 14 '19 at 07:50
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Sep 14 '19 at 07:52
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I am sorry , I didn't get ** -1 is not quadratic residue modulo p ** part – Aditya Singh Rathore Sep 14 '19 at 07:52
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That is only another formulation of "$x^2\equiv -1\mod p$ has no solution". – Peter Sep 14 '19 at 07:54
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Thank you ! got answer here link – Aditya Singh Rathore Sep 14 '19 at 07:56
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2If $p\equiv3\pmod4$ then $p-1$ is not divisible by four. Therefore there cannot be an element of order four in the group $\Bbb{Z}_p^*$ for that would violate Lagrange's theorem. – Jyrki Lahtonen Sep 14 '19 at 07:56