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I wanted you guys to tell me if my proof is correct:

By definition, $a^n ≡ a \bmod(n)$, if $n$ is Carmichael.

Assume $n$ is even. Then $n = 2m$, for some number $m$.

Then $a^{2m} ≡ a \bmod(2m)$. Since $2m$ is congruent to $\bmod(2m)$, $a^{2m} ≡ a^0 ≡ 1 ≡ a \bmod(2m)$.

Then, $a ≡ 1 \bmod(n)$ implies that $a ≡ 1, 1+n, 1+2n,...,1+(n-1)n$. But since $n$ is even, by assumption, the last term becomes $1+(2m-1)\cdot2m ≡ 1+2m$. But this term is odd for any $m$, hence there is a contradiction and $n$ is odd.

Is this correct?

Bill Dubuque
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user80979
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2 Answers2

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Hint: Let $a=n-1$.

Remark: There is no progress towards a proof in the OP.

André Nicolas
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To answer your question, no this is not correct. Your line "Since $2m$ is congruent to $\bmod(2m)$..." does not make sense. Replace $2m$ with $n$ and you will see that you can literally write $n$ is congruent to $\bmod(n)$, therefore $a^n ≡ 1 \bmod(n)$, which is not right.

Additionally, one detail you are missing is that $a^n ≡ a \bmod(n)$ is true for numbers between $1$ and $n$. So, writing $a ≡ 1, 1+n,...$ does not make sense either.

To prove that $n$ is odd, use the Korselt's criterion.

mcisse
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