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Let $p$ be a prime and let $\mathbb{Z}_p^* = \{1,2...p-1 \}$.

Let $a\in \mathbb{Z}_p^*$. Find $a^{-1}a =1$ for $a^{-1} \in \mathbb{Z}_p^*$.

So $a^{-1}a = 1\pmod{p} \iff a^{-1}a + pr =1$ , for some $r\in \mathbb{Z}$

Since $a$ and $p$ are coprime: $\gcd(a,p) =1$ and $\alpha a + \beta p =1$ has a solution for $\alpha , \beta \in \mathbb{Z}$ by Bezout's lemma.

Plugging in we get $\alpha a = 1 \pmod{p} \iff \alpha a = 1$, so $a=\alpha^{-1}$ with $\alpha = a^{-1}$.

Now $a= (a^{-1})^{-1} = a$

(Is this sufficient/enough?)

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  • $\begingroup$ This problem is related to this one $\endgroup$ – Juniven Mar 17 '17 at 12:46
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    $\begingroup$ Yes, the proof is correct. $\endgroup$ – freakish Mar 17 '17 at 15:02
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Your proof is correct, although the $a= (a^{-1})^{-1} = a$ at the end is not necessary.

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