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Trying to solve the modular equation $$x \equiv 58^{5745} \pmod{59}$$

I've seen a solution to this that involves using that $58 \equiv (-1) \pmod{59}$, and this in turn means that $58^{5745} \equiv (-1)\pmod{59}$, but I'm missing the rule that explains this step.

Can someone clarify?

Alec
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    @ArnaudD. No. OP has the solution, but is looking for an explanation. See Wikipedia: compatibility with exponentiation. This is provable by induction (or binomial theorem). – player3236 Nov 08 '20 at 15:48
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    $58\equiv -1\mod 59$ holds because of $59\mid 58+1$. Hence we can replace $58$ by $-1$. Since the exponent is odd, the result follows. – Peter Nov 08 '20 at 15:48
  • @player3236 - That's a great resource, thank you! – Alec Nov 08 '20 at 15:57

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Because $58 \equiv 59-1\equiv -1\pmod{59}$, $58^{5745}\equiv (-1)^{5745}\equiv (-1)\cdot ((-1)^{2})^{2871}\equiv \boxed{-1\pmod{59}}$

Joshua Wang
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