1
$\begingroup$

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:

$$n^{p-2} \equiv \frac{p+1}{n} \pmod p$$

$\endgroup$
4
$\begingroup$

$$n^{p-1}\equiv 1 \equiv p+1 \pmod p \implies n^{-1}n^{p-1}\equiv n^{-1}(p+1)\pmod p \iff n^{p-2}\equiv \frac {p+1}n$$

$\endgroup$
1
  • $\begingroup$ Very good, +1.. $\endgroup$ – Domenico Vuono Oct 24 '15 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.