$[(x-a_1)(x-a_2) \cdots (x-a_n)]^2 +1$ is irreducible over $\mathbb Q$ 
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.
 A: 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: $$g(x)=x^k+b_{k-1}x^{k - 1}+\cdots+b_0,\;h(x)=x^l+c_{l-1}x^{l-1}+\cdots+c_0\in \mathbb Z[x]  $$   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.
Conclusion
In both cases the supposed non-trivial factorizability of $f$ leads to a contradiction and thus $f$ is actually irreducible.
