Show that $\dfrac{2^p}{p}$ has remainder of $2$ for any prime $p \geq 3$

Question

A bonus question on my last math exam I haven't been able to solve. Thanks for the help.

Answer 1

Hint: Fermat's Little Theorem.

If you cannot or don't wish to directly employ that, you can consider the binomial expansion of $2^p = (1+1)^p$.

Answer 2

Fermat's little theorem is probably the best way to go here. Try to avoid it and you'll probably waste some time reinventing the wheel.

If $\gcd(b, p) = 1$ and $p$ is prime, then $b^{p - 1} \equiv 1 \pmod p$. This is Fermat's little theorem, it doesn't need to be proven again, but you can certainly look up a proof of it if you don't want to take my word for it.

So, if $p$ is an odd prime, then obviously $\gcd(2, p) = 1$. This means that $2^{p - 1} \equiv 1 \pmod p$. Since $1 \times 2 = 2$, this means that $2^p \equiv 2 \pmod p$.