Fermat's Little Theorem tells us for a prime $p$ if $\gcd(p,n)=1$, then:
$$n^{p-1} \equiv 1 \pmod p$$
If $n | (p+1)$, does it necessarily follow that:
$$n^{p-2} \equiv \frac{p+1}{n} \pmod p$$
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$$n^{p-1}\equiv 1 \equiv p+1 \pmod p \implies n^{-1}n^{p-1}\equiv n^{-1}(p+1)\pmod p \iff n^{p-2}\equiv \frac {p+1}n$$
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$\begingroup$ Very good, +1.. $\endgroup$Domenico Vuono– Domenico Vuono2015-10-24 15:38:39 +00:00Commented Oct 24, 2015 at 15:38