I'm trying to solve the following exercise:
Show that for $\alpha,\beta\geq 3$, the polynomial $f = X(X-3)(X-\alpha)(X-\beta) + 1\in\mathbb Z[X]$ is irreducible.
It is straightforward to check that the polynomial $f$ doesn't have any rational roots. So the only remaining possibility is a decomposition into $2$ factors of degree $2$. Here, I don't know how to proceed.
The factorization $$ X(X-1)(X-2)(X-3) + 1 = (X^2 - 3X + 1)^2 $$ shows that it won't be possible to get a contradiction via the reduction modulo any number $n$.
Based on the comments, I was able to solve the problem.
First, the case $\alpha = \beta = 3$ is routinely shown to be irreducible. (For example by looking at the image mod $2$, which is irreducible.)
Now we may assume $\alpha \geq 4$.
It is straightforward to check that $f$ doesn't have a rational root. So $f$ does not have a linear factor in $\mathbb Z[X]$, and thus the only remaining possibility is $f = gh$ with $g,h\in\mathbb Z[X]$ of degree $2$. We get $1 = f(0) = g(0)h(0)$, so either $g(0) = h(0) = 1$ or $g(0) = h(0) = -1$. Similarly, $g(3) = h(3)$ and $g(\alpha) = h(\alpha)$. So the polynomials $g$ and $h$ coincide in three different positions. With $\deg(g) = \deg(h) = 2$, this forces $g = h$. So $f = g^2$ is a perfect square.
However, from $\alpha,\beta \geq 3$ we get $f(1) \leq 1\cdot (-2)^3 + 1 = -7$. Contradiction.