Let $P\in{\mathbb Q}[X]$ be a non-constant polynomial. Consider the polynomial $Q(X,Y)=P(X)-P(Y)+1$.
- Is $Q$ always irreducible in ${\mathbb Q}[X,Y]$ ?
- Is $Q(X,y)$ always irreducible in ${\mathbb Q}[X]$ for large enough $y\in{\mathbb N}$ ?
My toughts : the main property of $Q$ is that $Q(X,X)=1$. So if $R$ is a divisor of $Q$, then $R(X,X)$ must be a constant.
Those questions ocurred to me while thinking about a recent question.
The answer to question (1) is yes. Suppose we have a factorization $Q=RS$ where $R$ and $S$ are nonconstant. Let $R_0$ and $S_0$ be the top-degree homogeneous parts of $R$ and $S$; then $R_0S_0=a(X^n-Y^n)$ where $n=\deg P$ and $a$ is the leading coefficent of $P$. Now observe that $X^n-Y^n=(X-Y)(X^{n-1}+YX^{n-2}+\dots+Y^{n-1})$. One of $R_0$ and $S_0$ contains the factor $X-Y$, so the other one (let's say $R_0$) is a factor of the polynomial $T(X,Y)=X^{n-1}+YX^{n-2}+\dots+Y^{n-1}$. Now note that in order for $R(X,X)$ to be a constant, $R_0(X,X)$ must be $0$ (since $R_0(X,X)$ is a positive degree homogeneous part of $R(X,X)$). But $R_0(X,X)\neq 0$ since it is a factor of $T(X,X)=nX^{n-1}$. This is a contradiction.
More generally, the same argument works over any field in which $n\neq 0$ (so that $T(X,X)\neq 0$). For a counterexample in positive characteristic, note that $X^p-Y^p+1$ factors as $(X-Y+1)^p$ over $\mathbb{F}_p$.