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I want to prove that there is an infinite number of primes $p$ for which the congruence $(p-3)x^{p-3}\equiv 1\pmod p$ has a solution. Using Dirichlet's theorem, I got that there is an infinite number of primes of the form $p=6k+1$, and because $(1/p)=1$, we get that there is an $x$ with $x^2\equiv 1\pmod p$ so that $x^{2(3k-1)}\equiv 1\pmod p$, whice gives $x^{p-3}\equiv 1\pmod p$.

How do I proceed?

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    Why write it this way? This is equivalent to $x^2\equiv -3\pmod p$, which clearly is solvable for infinitely many $p$. – lulu Jun 21 '24 at 14:17
  • $p-3\equiv -3\pmod p$ and $x^{p-3}\equiv x^px^{-3}\equiv x^{-2}\pmod p$. And note that $x$ can't be $0$. – lulu Jun 21 '24 at 14:44
  • $-3x^{p-3}\equiv1\iff-3\equiv x^2$ using Fermat’s little theorem $x^{p-1}\equiv1$ – J. W. Tanner Jun 21 '24 at 14:44
  • Scaling by $,x^2\not\equiv 0,$ shows $,1\equiv -3x^{p-3}\iff x^2\equiv -3x^{p-1}\equiv -3.,$ But it is easy to show $-3$ is a square mod $p$ for infinitely many $p,,$ e.g. see linked dupes. Beware that the accepted answer was incorrect. – Bill Dubuque Jun 21 '24 at 18:52

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We can first simplify the congruence equation (assume $p>3$):

$$(p-3)x^{p-3}\equiv -3x^{p-3} \equiv -3 x^{-2} \equiv 1 \Leftrightarrow x^2+3\equiv 0$$

where we have used $x^{p-1}\equiv 1$, as $x$ obviously cannot be $0$.

By Chebotarev's density theorem, $x^2+3$ splits mod $p$ for infinitely many $p$. Or more elementarily, by quadratic reciprocity,

$$\big(\frac{-3}{p}\big)=\big(\frac{-1}{p}\big)\big(\frac{3}{p}\big)=(-1)^{(p-1)/2}(-1)^{(p-1)/2}\big(\frac{p}{3}\big)=\big(\frac{p}{3})$$

Therefore $\big(\frac{-3}{p}\big)=1$ iff $p\equiv 1\mod 3$. By Dirichlet's theorem, there are definitely infinitely many such $p$.

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