Two questions about $a^{2^{n-1}} \mod {2^{n}}$ and Fermat's Little Theorem

Question

This is a question that comes in two parts.

First, prove that if $a^{2^{n-1}}\not\equiv1 \pmod {2^{n}}$, then $a$ is even.

Second, prove that for any odd prime $p$, if $a$ is odd and $a^{2^{n-1}(p-1)}\not\equiv1 \pmod 2^{n}p$, then $p|a$.

For the first one, I realized I can try proving the contrapositive, and then use induction. Then what should I do next? Fermat's Little Theorem?

For the second one, can you split the modulus into $2^{n}$ and $p$? I think this might be a good method. Answers and ideas are welcome!

Answer 1

For the first question, use https://mathworld.wolfram.com/EulersTotientTheorem.html or https://mathworld.wolfram.com/CarmichaelFunction.html

and https://en.m.wikipedia.org/wiki/Proof_by_contradiction

For the second, if $p\nmid a\iff(a,p)=1$

$$a^{p-1}\equiv 1\pmod p$$

If $(a,2)=1,$ using Carmichael or Totient function,

$$a^{2^{n-1}}\equiv1\pmod{2^n}$$

$$\implies a^{[p-1,2^{n-1}]}\equiv1\pmod{[p,2^{n-1}]}$$