I'm having trouble with the following exercise:
Let $k$ be an algebraically closed field with characteristic $\neq 2$, and let $f \in k[X]$. Show that $f-1$, $f$ and $f+1$ are all square only if $f \in k$.
Here's what I have so far: suppose $f = g^2$ for some $g \in k[X] \setminus k$. Then $f - 1 = g^2 - 1 = (g+1)(g-1)$. Since $k$ is a field of characteristic other than 2, $(g+1)$ and $(g-1)$ have completely disjoint sets of zeroes. This means that, if $f-1$ is square, then so are $g-1$ and $g+1$.
I think proving by induction on the polynomial's degree would be the best bet here, and it seems I somehow have to connect $f+1$ being square with $g$ also being a square (so I can apply the induction hypothesis). What am I missing?