# $[(x-a_1)(x-a_2) \cdots (x-a_n)]^2 +1$ 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.

• 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, 2013 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. Mar 1, 2013 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, 2013 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. Mar 1, 2013 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, 2013 at 7:47

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.

• +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 :-) Mar 1, 2013 at 9:16
• Dear @Jyrki, I'm very happy if I helped you stop worrying :-) Mar 1, 2013 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. Jun 19, 2015 at 10:56