In one proof I'm trying to figure out, I've arrived at a part where I am thinking about whether $x≡1$ and $y≡1$ imply $x^ay^b≡1$ for any real integers a and b. I know that in general, if $u≡v$ and $c≡d$, then $ub≡vd$. But what if one of $a$ and $b$ is negative? Let's say that $a≡1$. Can you prove that $a^k≡1$ for any integer k, even if k could be negative?

Thank you in advance.

  • 1
    $\begingroup$ In modular arithmetic, $x^{-1} \pmod{n}$ means the residue $y\pmod{n}$ so that $xy \equiv 1\pmod{n}$, if it exists. $\endgroup$ – Jair Taylor Mar 12 at 6:50
  • $\begingroup$ In particular, if one wants to prove that $a^{-j}\equiv1\pmod m$, one first needs to know the definition of the notation $a^{-j}\pmod m$. The proof must start from that! $\endgroup$ – Greg Martin Mar 12 at 7:18
  • $\begingroup$ If $\,x^n\equiv 1$ for an integer $n\ge 1$ then $\,xx^{n-1}\equiv 1\,$ so $\,x^{-1}\equiv x^{n-1}\,$ so $\,x^{-k}\equiv x^{(n-1)k}\ $ $\endgroup$ – Bill Dubuque Mar 14 at 3:10

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