Proposition: Let $2^i$ be the highest power of 2 dividing m, let a be odd and assume that $x^{m} \equiv a\space (mod\space 2 ^ {2i+1}$ ) is solvable. Then $\forall$ $j \geq$ 2i + 1, $x^{m} \equiv a (mod\space 2^{j} )$ is solvable.
I am seriously struggling to understand why the proof below does not work for p = 2:
Proposition: Let p be an odd prime, $p \nmid a$, $p \nmid$ m. Then if $x^{m} \equiv a $(mod p) is solvable, so is $x^{m} \equiv a (mod\space p^{j} )$, for$ j \geq 1$.
Proof: By induction. If j = 1, it is trivially true. Suppose it is true for j > 1. Let $x_{j}$ be such a solution, and choose $\beta$ to be the unique integer such that $mx_{j}^{m-1} \beta \equiv (a - x_{j}^{m} )/p^{j}$ (mod p). This is possible as $(a - x_{j}^{m} )/p^{j} \in \mathbb{Z}$ by assumption, and $p \nmid mx_{j}^{m-1}$ . Then $x_{j+1} := x_{j} + \beta p^{j}$ is such that:
$ \; \begin{array}{ccc} x_{j+1}^{m} & = & \sum_{k=1}^{m}\dbinom{m}{k}x_{j}^{k}(\beta p^{j})^{m-k}\end{array}$
and by taking congruences we see that $x_{j+1}^{m} \equiv x_{j}^{m} + mx_{j}^{m-1} p^{j} \beta (mod \space p^{j+1} )$. By our choice of $\beta$ , $mx_{j}^{m-1} p^{j} \beta \equiv a - x_{j}^{m} (mod \space p^{j+1} )$, showing $x_{j+1}$ is a solution to the equation $x^{m} \equiv a (mod\space p^{j+1} )$, hence the result.$ \square$
It would be much help if someone could explain what goes wrong in this proof when p = 2, I kinda need this result for the essay I have been writing for a year! Cheers