I am asked the following:
Show that the number of integer solutions to $y^p=x^2+2$ for any odd prime $p$ is at most $p-1$.
I checked that for $y^p=x^2+2$, the same method for $y^3=x^2+2$ works and I get that $x+\sqrt{-2}=(a+b\sqrt{-2})^p$ for some integer $a$ and $b$. I then expand RHS with Binomial Theorem and get $$\sum_{k=0}^p {p\choose k}a^k (b\sqrt{-2})^{p-k}$$
Since $(\sqrt{-2})^n$ is real whenever $n$ is even, equate real and imaginary parts and get $$x=\sum_{k=0, k \mbox{ odd}}^p {p\choose k}a^k (b\sqrt{-2})^{p-k}$$ and $$\sqrt{-2}=\sum_{k=0, k \mbox{ even}}^{p-1} {p\choose k}a^k (b\sqrt{-2})^{p-k}$$ Divide through by $\sqrt{-2}$ and take $b$ out to get $$1=b\left(\sum_{k=0, k \mbox{ even}}^{p-1} {p\choose k}a^k (b\sqrt{-2})^{p-k-1}\right)$$
So we must have $b=1$ or $b=-1$. Also, the horrible sum inside the bracket is a polynomial of degree $p-1$, so for each choice of $b$ there can be at most $p-1$ (integer) choices of $a$.
Now here is the problem. Because there are two choices of $b$, there are at most $2(p-1)$ choices of $(a,b)$ that satisfy the last equality. Thus there are at most $2(p-1)$ choices of $x$. But I can't seem to find a way to lower this upper bound to just $p-1$. I am feeling so close! Any help will be appreciated!
