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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 $n$.

azimut
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  • In fact your polynomial is reducible only in the case you mentioned, that is, the numbers are consecutive. See http://math.stackexchange.com/questions/579971/proof-that-a-polynomial-is-irreducible-for-all-n-ne-4 or http://www.jstor.org/stable/pdf/2970062.pdf?acceptTC=true – user26857 May 25 '15 at 22:57
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    @user26857: Thank you for the pointer. So I guess it remains to show that the given polynomial is not a perfect square in $\mathbb Z[X]$. – azimut May 25 '15 at 23:28
  • Are $\alpha,\beta$ assumed to be distinct? Only then the argument in the link works. – Martin Brandenburg May 25 '15 at 23:49
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    The polynomial is not a square because it's negative for $x=1$. – user225222 May 26 '15 at 00:22
  • @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. – user225222 May 26 '15 at 00:30
  • I think u can easily weaken the conditions on $\alpha,\beta$. So ma bee there is another trick?! – user225222 May 26 '15 at 00:37
  • @MartinBrandenburg: No, they are not. But given the condition $\alpha,\beta \geq 3$, the only problematic case is $\alpha = \beta = 3$, which is easily excluded looking at the reduction modulo $2$. Otherwise, $g$ and $h$ coincide in at least three different values. By $\deg(g) = \deg(h) = 2$, this implies $g = h$. – azimut May 26 '15 at 07:43
  • @user225222: This is the exact version which I found as an exercise. Yes, I agree that this question should be doable with less restrictive conditions. Maybe it is possible to characterize all irreducible polynomials $(x-\alpha)(x-\beta)(x-\gamma)(x-\delta) + 1\in\mathbb Z[x]$ with $\alpha,\beta,\gamma,\delta\in\mathbb Z$? – azimut May 26 '15 at 07:47

1 Answers1

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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.

user26857
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azimut
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