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I'm currently working in the following Fermat's Little theorem exercise:

Let $p$ be an odd prime and let $i ≥ 0$. Show that $2^i$ $\not\equiv$ $2^{p+i}$ $\mod p$.

My thoughts are the following:

By the corollary of Fermat's little theorem if $2^i$ $\equiv$ $2^{p+i}$ must be true that:

$$i\equiv{p+i} \ mod (p-1)$$

Or another interpretation of the congruence:

$$(p-1)k | i-(p+i) $$

Then

$$(p-1)k | i-p-i $$ $$(p-1)k | -p $$

Which is impossible because $p-1$ is not a prime number.

I'm not sure about that to be a good proof for the requested exercise as there is this remark on the textbook:

"This shows that having the exponents congruent mod p does not yield an overall congruence mod p."

Any comment or suggestion will be really appreciated.

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Well, by Fermat's little theorem, $2^p\cong2\pmod p$. So, $2\cdot2^i\cong2^{p+i}\cong2^i\pmod p\implies 2^i\cong0\pmod p$, a contradiction.

This result does indeed show that exponents congruent $\pmod p$ doesn't necessarily mean we have congruence, since $p+i\cong i\pmod p$.

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