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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 $n$.

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  • $\begingroup$ In fact your polynomial is reducible only in the case you mentioned, that is, the numbers are consecutive. See math.stackexchange.com/questions/579971/… or jstor.org/stable/pdf/2970062.pdf?acceptTC=true $\endgroup$ – user26857 May 25 '15 at 22:57
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    $\begingroup$ @user26857: Thank you for the pointer. So I guess it remains to show that the given polynomial is not a perfect square in $\mathbb Z[X]$. $\endgroup$ – azimut May 25 '15 at 23:28
  • $\begingroup$ Are $\alpha,\beta$ assumed to be distinct? Only then the argument in the link works. $\endgroup$ – Martin Brandenburg May 25 '15 at 23:49
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    $\begingroup$ The polynomial is not a square because it's negative for $x=1$. $\endgroup$ – user225222 May 26 '15 at 0:22
  • $\begingroup$ @MartinBrandenburg, no need that $\alpha \neq \beta$. U can adapt the proof in math.stackexchange.com/questions/579971/…assuming that $g,h$ are of degree 2. $\endgroup$ – user225222 May 26 '15 at 0:30
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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.

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