Problem: Use fermat's little theorem to show that for $p$ (a prime number), every integer $a$ with $a\not \equiv 0 \pmod{p}$ has a reciprocal modulo $p$.

The definition of a reciprocal is: for $a,b,n \in \mathbb Z$ with $n \ge 1 $ we call $b$, a reciprocal of $a$ if $ab \equiv 1 \pmod{n}$.

Can someone help me with the proof of this I am not even sure where to start.

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    $\begingroup$ $a^{p-2}.a=a^{p-1}=1 mod p$ $\endgroup$ – ali Nov 3 '16 at 6:12

The proof is immediate if we consider $$a^{p-1}\equiv 1 \pmod{p}$$ Then $a^{p-1}=a\cdot a^{p-2}$. The reciprocal modulo $p$ of $a$ is $a^{p-2}$.

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