I know that for $m \in \mathbb{Z}$ then $a$ has an inverse modulo $m$ if and only if $\gcd(a,m) = 1$ meaning $a$ and $m$ are relatively prime and their greatest common divisor is 1

Prove that for all integers $n>1$, the number $n-1$ has an inverse modulo $n$. Thank you


closed as off-topic by Hurkyl, user223391, Ivo Terek, José Carlos Santos, Dando18 Oct 5 '17 at 0:02

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  • 1
    $\begingroup$ I feel like I already answered this this morning. $\endgroup$ – Randall Oct 4 '17 at 1:50
  • 3
    $\begingroup$ Hint, $n-1\equiv -1$ modulo $n$. What's a plausible (multiplicative) inverse for $-1$? $\endgroup$ – Mark Oct 4 '17 at 1:57
  • $\begingroup$ what factors can $n$ and $n-1$ share? $\endgroup$ – XRBtoTheMOON Oct 4 '17 at 2:09

From the question you wrote, it seems like you already know a potential plan to solve this problem: simply prove $\gcd(n-1, n) = 1$.

This works. Do it. (you should have tried it even without the reassurance that it works!)


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