We call a two-variable polynomial $P(x, y)$ secretly one-variable, if there exist polynomials $Q(x)$ and $R(x, y)$ such that $\deg(Q) \ge 2$ and $P(x, y) = Q(R(x, y))$ (e.g. $x^2 + 1$ and $x^2y^2 +1$ are secretly one-variable, but $xy + 1$ is not). Prove or disprove the following statement: If $P(x, y)$ is a polynomial such that both $P(x, y)$ and $P(x, y) + 1$ can be written as the product of two non-constant polynomials, then $P$ is secretly one-variable. Note: All polynomials are assumed to have real coefficients.
Problem
Source: Baltic Way 2022, Problem 3
Tags: algebra, polynomial
Tintarn
03.03.2023 14:02
This problem was proposed by me. It arose from a lunch discussion going something like this (this story might actually be more interesting than the problem itself ...): Someone raised the question of whether there is a polynomial $P$ such that both $P(x)$ and $P(x)+1$ are reducible. Quickly, we found $P(x)=x^2-1$ as a solution. We then asked whether we can classify all the solutions, but this turns out to be quite complicated. We then went on to ask what happens if we ask for polynomials in several variables. Soon we realized that this question was ill-posed: If we have a solution to the one-variable problem, then we can always still just take $P(x,y)=x^2-1$ as a solution (i.e. just let $P$ not depend on the second variable). But even if we force $P(x,y)$ to actually depend on both variables, there are a lot of trivial solutions like $P(x,y)=x^2y^2-1$ or more generally $P(x,y)=Q(R(x,y))$ where $Q(x)$ is a solution to the one-variable problem and $R(x,y)$ is arbitrary. We then asked: Do all solutions for the two-variable problem arise in this way from the one-variable problem? And thus, the notion of "secretely one-variable" was born. The way we went on to solve the problem originally was like this: We realized that solutions to the problem correspond to matrices with polynomial entries and determinant $1$ since $\det \begin{pmatrix} a & b\\c & d \end{pmatrix}=ad-bc$ and so $ad-bc=1$ implies that we can take $p=bc$ and $p+1=ad$. For instance, the solution $P(x)=x^2-1$ corresponds to the matrix $\begin{pmatrix} x & x-1\\x+1 & x \end{pmatrix}$. (We are thus asking, in fancy terms, for the structure of the group $SL_2(\mathbb{R}[x,y])$.) But from this point of view, there is a natural way to produce new examples: Take two matrices and multiply them. Starting from the "secretely one-variable" solutions, we can therefore construct the new example \[\begin{pmatrix} x & x-1\\x+1 & x \end{pmatrix} \cdot \begin{pmatrix} y & y-1\\y+1 & y \end{pmatrix}=\begin{pmatrix} 2xy+y-x-1 & 2xy-x-y\\2xy+x+y & 2xy+x-y-1 \end{pmatrix}.\]And so we get the solution \[P(x,y)=4x^2y^2-x^2-y^2-2xy=(2xy+x+y)(2xy-x-y)\]with \[P(x,y)+1=4x^2y^2-x^2-y^2-2xy+1=(2xy+x-y-1)(2xy+y-x-1)\]and all that remains to be checked is that $P(x,y)$ is indeed not secretely one-variable which is a rather boring but not very difficult thing, given that there are not so many possibilities for the degree of $Q$ and $R$. (Of course the example in #2 is much simpler than this one.)