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?)

  • $\begingroup$ This problem is related to this one $\endgroup$ – Juniven Mar 17 '17 at 12:46
  • 1
    $\begingroup$ Yes, the proof is correct. $\endgroup$ – freakish Mar 17 '17 at 15:02

Your proof is correct, although the $a= (a^{-1})^{-1} = a$ at the end is not necessary.

| cite | improve this answer | |

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.