We know that if, $a\equiv b \pmod n$ then $$a^n\equiv b^n \pmod n$$ Is the converse true for odd $n$? The converse isn't true for even because $3^4 \equiv 1^4 \pmod 4$ but $3 \equiv 1 \pmod 4$ isn't true. If the converse is true for odd n then give me a proof of this. If the converse isn't true for all odd $n$ then what is the condition of $(a,b,n)$ so that the converse is true.
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the converse is not true for all composite odd $n$; e.g., $1^9\equiv4^9\bmod9$ – J. W. Tanner Apr 07 '20 at 02:30
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It's true for prime $n$ because $a^n\equiv a\pmod n$ for prime $n$ by Fermat's little theorem.
This is also true for Carmichael numbers, which are composite.
J. W. Tanner
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