Lemma: If there exists solutions in to with odd prime , then there exists a solution such that

Proof: Let denote the order of mod , an abitrary solution, we can represent all the solutions as with . Since because , therefore which means there exists a solution congruent to mod .

Direct Correlary: If there exists solutions in to with odd prime , then there exists such that or .

Main Proof:
Consider the values of .

If then the constant coefficient of the polynomial is also even which then contradicts the coprime condition.

If there exists odd prime such that , then is a solution to , by our direct correlary above there exists such that . However, a well-known property of integer polynomial tells us that $p\mid 2^{m'}-m'\mid f(2^{m'})-f(m')\Rightarrow (f(m'),f(2^m'))=(f(m'),f(2^{m'})-f(m'))=pk$ for some , which doesn't satisfy our intended condition.

These two conditions forces for all which means one of has an infinite number of solution which is impossible unless