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

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.

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 $$p+i\cong i\pmod p$$.