Let $f(x) = (x - a_1)(x - a_2)...(x - a_n) + 1$, where $a_1,a_2,a_3,\dots , a_n$ are distinct integers. Show that

  1. if $n$ is odd , then $f(x)$ is irreducible over $\mathbb{Z}$, i.e $f(x)$ cannot be factorized in the form $f(x) = p(x)q(x)$ where $p(x)$ and $q(x)$ are polynomials with integer coefficients and their degrees are less than the degree of $f(x)$ ( which is $n$ here).

  2. If $n$ is even, either $f(x)$ is irreducible over $\mathbb{Z}$ or is the square of a polynomials with integer coefficients

  • $\begingroup$ See here for a special case. I need to check to what extent the argument survives when we have any integers $a_i$ as opposed to $1,2,3,\ldots,n$. Looks like it might survive precisely in the required form, but.... $\endgroup$ – Jyrki Lahtonen Jul 12 '17 at 11:03
  • $\begingroup$ Yup, my answer from the linked thread works with the obvious modifications about the location of zeros in place. It comes to MooS's answer exactly. $\endgroup$ – Jyrki Lahtonen Jul 12 '17 at 11:09

If $f$ is reducible, say $f=pq$, we get that $p(a_i)q(a_i)=1$ for $1 \leq i \leq n$.

This shows that $p(a_i)=q(a_i)=\pm 1$ and one of those two choices has to occur at least $\lceil \frac{n}{2} \rceil$ times.

If $n$ is odd, this shows $\deg p$ and $\deg q$ are both greater than $\frac{n}{2}$, contradiction!

If $n$ is even, this shows that $\deg p=\deg q=\frac{n}{2}$ and $p-q$ has $n > \frac{n}{2}$ roots, thus $p=q$.

  • $\begingroup$ Of course one has to invoke the Gauß lemma to make sure that the factors have integer coefficients, i.e. $p(a_i)$ and $q(a_i)$ are really integers, but I thought that goes without saying. $\endgroup$ – MooS Jul 12 '17 at 11:12
  • $\begingroup$ Also this proof shows that in the even case, $f$ is actually a square of an irreducible polynomial. Because if $f=p^2$ for $p$ reducible - say $p=p_1p_2$, you could factor $f$ as $f=p_1(p_1p_2^2)$, which is not possible as seen in the proof. $\endgroup$ – MooS Jul 12 '17 at 11:15

$(1)$ At first let $a_i<a_{i+1}$ We can know possible solution is $x<a1$. At least $x\leq a_1-1$.


Since $f(x)$ is monotonically increasing at $x<a_1-1$, $f(a_1-1)$ is obviously negative number, then $f(x)$ has no integer solution at $x<a_1-1$.

$(2)$We assume $a_i<x_i<a_{i+1}$ is solution.


,which is not $0$ in the same way as previous discussion.

If $f(x)$ is square number at every integer,


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