I am trying to show the following statement: Let $k$ be a field (no assumptions about algebraic closure or similar). Let $c \in k$ be a constant, which is not $1$ or $0$. Let $p(x) \in k[x]$ be a nonconstant polynomial, then $q(x) := p(x)(p(x) - 1)(p(x) - c)$ is not a square in $k[x]$.
It seems to me that under the assumption of $k$ being algebraically closed this reduces to what the title states, by observing that the three factors are coprime and hence share no roots, so every root of $p(x)$, $p(x) - 1$, and $p(x) - c$ would have multiplicity $2$ if $q(x)$ was a square, hence they would all be squares.
- Does this reduction of the problem work when $k$ is not algebraically closed?
- If yes, how can one show the statement in the title?
- If no, how does one show the statement in the first paragraph without this reduction? Is it even true?
I was not able to find any counterexample to the statement of the first paragraph (I mostly looked at fields of characteristic $2$). It should be true as it is equivalent to an exercise from an algebraic geometry class I am currently taking, although that exercise could be lacking assumptions.
Assuming that $k$ is algebraically closed and its characteristic is distinct from $2$, it seems that you cannot even have two "consecutive squares"; that is, $P$ and $P-1$ cannot both be squares (unless they are constant polynomials). To see this, assume for a contradiction that this is possible, and let $P$ and $P-1$ are polynomials of the smallest possible (positive) degree which are both squares; say, $P=Q^2$ and $P-1=R^2$. Then $R^2=Q^2-1=(Q-1)(Q+1)$, and since $Q-1$ and $Q+1$ are coprime, they are both squares: $Q-1=U^2$, $Q+1=V^2$. It follows that $V^2-U^2=2$, and therefore $V^2/2$ and $V^2/2-1$ is yet another pair of "consecutive squares" (recall that $k$ is algebraically closed), while $\deg (V^2/2)=\deg Q=(\deg P)/2<\deg P$, a contradiction.