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Can we describe those prime numbers $p$ for which $p-1$ is a perfect square $\bmod p$.

For example, it is true for $p=2$ and $p=5$.

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  • $\begingroup$ You mean so that there is a number $a$ so that $a^2 \equiv (-1) \mod{p}$? Well, for $11$ it doesn't work ... $\endgroup$ – Matti P. Oct 25 '19 at 8:01
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    $\begingroup$ 1 mod 4, and 2 ${}$ $\endgroup$ – mathworker21 Oct 25 '19 at 8:01
  • $\begingroup$ @MattiP. Yes. It also doesn't work for $p=7$. But it does work for $p=13$ $\endgroup$ – Tusif Ahmed Oct 25 '19 at 8:08
  • $\begingroup$ So the answer is no. $\endgroup$ – Matti P. Oct 25 '19 at 8:11
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$2$ works, so suppose $p$ is odd. If $a^2 \equiv -1 \pmod{p}$, then $a^4 \equiv 1 \pmod{p}$, so the order of $a$ mod $p$ divides $4$. Since $a \not \equiv 1$ and $a^2 \not \equiv 1$ (since $-1 \not \equiv 1$, since $p > 2$), we must have $ord_p(a) = 4$. Since $a^{p-1} \equiv 1 \pmod{p}$ (Fermat's little theorem), we must have $4 \mid p-1$, i.e., $p \equiv 1 \pmod{4}$.

Now suppose $p \equiv 1 \pmod{4}$. Then $x^4-1$ divides the polynomial $x^{p-1}-1$. Since $x^{p-1}-1$ has exactly $p-1$ roots in $\mathbb{Z}_p$, and since $\frac{x^{p-1}-1}{x^4-1}$ has at most $p-5$ roots (since $\frac{x^{p-1}-1}{x^4-1}$ has degree $p-5$), it must be that $x^4-1$ has exactly $4$ roots. Since $x^2-1$ has exactly $2$ roots, it must be that $x^2+1$ has exactly $2$ roots. So it in particular has a root, which is what we want.

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Check this link for more information

https://en.m.wikipedia.org/wiki/Legendre_symbol

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    $\begingroup$ This should be a comment. Either that, or edit your answer to add in the essential parts of the wikipedia page to make the answer self contained. $\endgroup$ – YiFan Oct 25 '19 at 9:37
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    $\begingroup$ @YiFan it seems like you're defining "answer" to be "self-contained answer" $\endgroup$ – mathworker21 Oct 25 '19 at 10:07
  • $\begingroup$ @marh: as is usual on this site, as contrasted to the usual practice on, say, MO. $\endgroup$ – YiFan Oct 25 '19 at 10:17

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