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Find all numbers that are their own multiplicative inverse in $mod$ $p$ where $p$ is prime.

I recall that when $p$ is prime, all integers from 1 to the modulus minus 1, so all numbers from $1$ to $p-1$, have multiplicative inverses in mod $p$. So, the numbers that are their own multiplicative inverse would be $1$ and $1-p$. Can someone please explain why?

Keith Axelrod
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  • We have $a \equiv a^{-1} \mod p$. If you multiply both sides by $a$ we get $a^2 \equiv 1 \mod p$. This solves as $(a + 1)(a - 1) \equiv 0 \mod p$. – orlp Jul 19 '17 at 22:40
  • @orlp: And then use fact that p is prime to argue $(a+1)(a-1)=0$ has only two solutions $mod p $ , since $\mathbb Z_p$ is a field, so an integral domain. – gary Jul 19 '17 at 23:09
  • @orlp how did you infer $a≡a^{−1}\pmod{p}$? – mehsheenman Jan 28 '24 at 04:18
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    @mehsheenman That's the premise of the question. – orlp Jan 28 '24 at 17:13

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You want to solve $x^2=1$ i.e. $(x-1)(x+1)=0$, which for a prime modulus gives $x=\pm 1$.

J.G.
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