Let $p_1, p_2, \cdots, p_n$ be distinct integers and let $f(x)$ be the polynomial of degree $n$ given by $$f(x) = (x - p_1)(x - p_2)\cdots (x -p_n)$$ Prove that the polynomial $g(x) = (f(x))^2 + 1$ cannot be expressed as the product of two non-constant polynomials with integral coefficients.
We prove this by contradition. Suppose $g(x)=h(x)k(x)$ where $h(x)$ and $k(x)$ are non-constant polynomials with integer coefficients.
Because $g(x) > 0$ holds for all $x$, $h(x)$ and $k(x)$ have no real roots and thus must not change signs for any $x$. Without loss of generality, let's assume $h(x)$ and $k(x)$ are both positive.
Additionally, since $g(p_i)=1$ for all $i=1,\cdots, n$, we must have $h(p_i)=k(p_i)=1$ as well, implying both $h(x)$ and $k(x)$ have a degree of $n$ due to the fact that $p_i$ are $n$ distinct numbers. Hence, we can assume $$\begin{align}h(x)=1+a(x-p_1)(x-p_2)\cdots(x-p_n)\\k(x)=1+b(x-p_1)(x-p_2)\cdots(x-p_n)\end{align}$$
where $a$ and $b$ are to-be-determined integer coefficients. It follows that on one hand $$g(x)=(f(x))^2 + 1=(x - p_1)^2(x - p_2)^2\cdots (x -p_n)^2+1$$
and on the other hand $$\begin{align}g(x)=h(x)g(x)&=\left(1+a(x-p_1)(x-p_2)\cdots(x-p_n)\right)\left(1+b(x-p_1)(x-p_2)\cdots(x-p_n)\right)\\&=1+(a+b)(x-p_1)(x-p_2)\cdots(x-p_n) + ab((x-p_1)^2(x-p_2)^2\cdots(x-p_n)^2\end{align}$$
Hence, we must have $$\begin{cases} a+b&=0\\ab&=1\end{cases}$$
However, this is insolvable in integers.