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Suppose that $a_1,a_2, \cdots, a_n$ are $n$ different integers. Then $[(x-a_1)(x-a_2) \cdots (x-a_n)]^2 +1$ is irreducible over $\mathbb Q$.

I've no idea why it is true. Thanks very much.

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It is a conjecture that if $g(x)$ is monic with $m$ distinct integer roots and $f(x) = x^{2^n} + 1$, then $f \circ g(x)$ is irreducible over $\mathbb{Q}$ with exceptions $n = 0$, $m \leq 4$. I know that a proof, for $n = 1$, is contained in Pólya and Szegő's book that can be found here. – J.H. Mar 1 '13 at 7:11
If the given polynomial is a product $f(x)g(x), f,g\in\mathbb{Z}[x]$, then we must have $(x-a_1)(x-a_2)\cdots (x-a_n)\mid f(x)-g(x)$, but I can't take that anywhere. – Jyrki Lahtonen Mar 1 '13 at 7:29
@JyrkiLahtonen if $f(x)$ is reducible over $\mathbb Q$ then $f(x)$ is reducible over $\mathbb Z$ if i take the contra-positive of the above statement which imples $f(x)$ is irreducible over $\mathbb Z$ then $f(x)$ is irreducible over $\mathbb Q$ is int this helpful? – jim Mar 1 '13 at 7:34
@jim: I already used that bit to get to this point. Sorry about not making it clear that this applies to factors in $\mathbb{Q}[x]$ (normalized to be monic) as well. – Jyrki Lahtonen Mar 1 '13 at 7:37
To add to my comment above, this is conjecture due to Schur. More information can be found here. In particular, Brauer et al. provided a proof for the cases $n = 1,2$. – J.H. Mar 1 '13 at 7:47
up vote 17 down vote accepted

This is an old (1909) result of Issai Schur which can be found in Pólya-Szegő , on page 133 of Volume 1 . Here is the proof of irreducibility:

Let $f(x)=(x-a_1)^2(x-a_2)^2 \cdots (x-a_n)^2 +1$ be your polynomial and suppose it factors non-trivially as $f(x)=g(x)h(x)$ over $\mathbb Q$.
By Gauss's lemma we may assume that $g,h$ are monic with integral coefficients: $$g(x)=x^k+b_{k-1}+\cdots+b_0,\;h(x)=x^l+c_{l-1}x^{l-1}+\cdots+c_0\in \mathbb Z[x] $$ Notice that the polynomial functions functions $g,h$ satisfy $g(r),h(r)\gt 0$ for $r\in \mathbb R$ and that $g(a_i)=h(a_i)=1$ .
We may assume $k\leq l $, so that $k\leq n$ since $k+l=2n$, and then we distinguish two cases:

Case 1: $k\lt l$
Then the polynomial $g$ takes the value $1$ for the $n$ distinct values $a_1,\cdots a_n$ and, because it has degree $k\lt n$, that polynomial is the constant $g=1$ and the factorization $f=gh$ is trivial, contrary to our assumption .

Case 2: $k=l=n$
Then $g-h$ is a polynomial of degree $\lt n$ vanishing at the $n$ numbers $a_i$, so that $g-h=0$ and $g=h$.
Hence the assumed factorization $f=gh$ becomes $f(x)=(x-a_1)^2(x-a_2)^2 \cdots (x-a_n)^2+1 =g(x)^2$ and we get $ 1= g(x)^2-(x-a_1)^2(x-a_2)^2 \cdots (x-a_n)^2$ so that $$1=[g(x)+(x-a_1)(x-a_2) \cdots (x-a_n)][g(x)-(x-a_1)(x-a_2) \cdots (x-a_n)]$$ This is a clearly absurd factorization of $1$ into positive degree polynomials.

In both cases the supposed non-trivial factorizability of $f$ leads to a contradiction and thus $f$ is actually irreducible.

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+1: The bit that I seem to have missed was that $f$ has no real zeros. Consequently neither do $g$ and $h$, and I can stop worrying about the possibility that $g(a_i)$, $1\le i\le n$, are a mixture of $\pm1$:s :-) – Jyrki Lahtonen Mar 1 '13 at 9:16
Dear @Jyrki, I'm very happy if I helped you stop worrying :-) – Georges Elencwajg Mar 1 '13 at 9:25
Nice answer! An alternative way to render case 2 absurd is to note that $g+h$ is a polynomial of degree $n>0$, whose derivative, which is of degree $n-1$, has at least $n$ zeros (use that $f'(a_i)=0$ and the product rule). But this is clearly impossible. – Nils Matthes Jun 19 '15 at 10:56

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