Basically while doing exercise (2) on Pg.6 of the book "A Conversational Introduction to Algebraic Number Theory by Paul Pollack", I had a number of doubts. I will state the full problem to avoid confusion :
Problem : Let $p$ be an odd prime. In this exercise, we outline a proof that there are no nonzero integer solutions to $y^2 = x^p -1$. Seeking a contradiction, suppose that $(x,y)$ is an integer solution with $xy \neq 0$.
(a) Show that $x$ is odd and $y$ is even.
(b) By comparing imaginary parts in the equation $y+i = (a+bi)^p$, prove that $\sum_{k \geq 0} {p \choose 2k} (-a^2)^k = \pm 1$. By looking modulo $4$, show that that $+$ sign holds.
(c) Let $v$ be the exponent of $2$ appearing in the prime factorisation of ${p \choose 2}a^2$. Show that
$$2^{v+1} \mid {p \choose 2k}(-a^2)^k$$ for every integer $k >1$.
I am clear with (a).
In (b), expanding $(a+bi)^p$ using binomial, I am getting $\sum_{k \geq 0} {p \choose 2k} (-a^2)^k b^{p-2k} = \pm 1$, instead of the equation in (b) of the problem. I highly contend this is just a printing mistake. In other part of (b), I think we have to look at $\sum_{k \geq 0} {p \choose 2k} (-a^2)^k = \pm 1$ modulo 4, but I don't see how it gives us that $+$ sign holds. If we had only $a^2$'s in the equation, then I would have thought about this, but here we have coefficients with this square and moreover a sum.
In (c), I thought of induction on $k$ first, got stuck in proving the base step only. Then, I saw that from ${p \choose 2k}$ we can extract ${p \choose 2}$. So, we already have $2^{v} \mid {p \choose 2k}(-a^2)^k$, I don't know how to get an extra factor of $2$.
Please help me with these doubts, and thanks a lot in advance!
Yes, you have $$\sum_k \binom{p}{2k}(-a)^kb^{p-2k}=\pm1.$$ But $b$ is a factor of the LHS, and so of $\pm1$. Thus $b=1$ or $-1$ and all the $b^{p-2k}=b=1$ or $-1$. Using this gives Pollack's formula.