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I think that I got the answer, but am not quite sure whether or not it is correct.

Prove that $a^{25}$ mod 65 = $a$ mod 65 given that $a \in \mathbb{Z}$

I started off by using the Chinese Remainder Theorem (65 = 5 $\cdot$ 13 and gcd(5, 13) = 1) in order to get the following two equations:

$a^{25}$ mod 5 = $a$ mod 5

and

$a^{25}$ mod 13 = $a$ mod 13

After that, I used the theorem that states that:

$a^e \equiv a^{e \textrm{ mod } \phi(n)} \textrm{ mod } n$ if gcd(a, n) = 1

$\phi(5) = 4$ and $\phi(13) = 12$

Applying said theorem to the two equations above:

$a^{25 \textrm{ mod } 4}$ mod 5 = $a^1$ mod 5 = $a$ mod 5

and

$a^{25 \textrm{ mod } 12}$ mod 13 = $a^1$ mod 13 = $a$ mod 13

I know that the theorem states that gcd(a, n) = 1 which isn't necessarily the case, that's why I wasn't sure whether my solution was correct.

mrMoonpenguin
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1 Answers1

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For prime $n$, we have $a^n\equiv a\pmod n$. This doesn't require any sort of assumption on the relationship between $a$ and $n$ (we can deduce it from your result for $\gcd(a,n)=1$, and for $\gcd(a,n)=n$, well, $0^n\equiv 0$ is obvious).

Modulo $5$ we get $$a^{25}=(a^5)^5\equiv a^5\equiv a$$Modulo $13$ we get $$a^{25}=a^{13}\cdot a^{12}\equiv a\cdot a^{12}\equiv a$$

Arthur
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    Good answer. And to emphasize for $a \equiv 0$ mod 5 it follws that $a^{25} \equiv 0$ mod 5 $\equiv$ $a$ mod 5.... – Mike Apr 30 '19 at 20:37
  • Right, thanks! I assume that this is a variation of Fermat's little theorem? – mrMoonpenguin Apr 30 '19 at 20:41
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    @mrMoonpenguin Yes, it is a special case of a generalization of Fermat & Euler's theorem - see my answer in the linked dupe and see the 2nd dupe thread for other methods (applied to a similar problem). – Bill Dubuque Apr 30 '19 at 21:22