Show that if $f(x)=15+\prod_{j=1}^{14}(x-j)^2$, then $f(x)$ is irreducible over $\mathbb{Q}$.
My thought: I have seen and known how to deal with $f(x)=1+\prod_{j=1}^{14}(x-a_j)^2$, in which $a_j$ is pairwise different integers. I think the difficulty is caused by the difference between $1$ and $15$. $15$ has two kinds of factorization $3\times5$ and $1\times15$ while $1$ has only one.
For the second case, supposed that $f(x)=g(x)h(x)$ in $\Bbb{Z}[x]$. W.l.o.g. $g(x)$ and $h(x)$ are monic and $\deg(g)<=14$. Firstly notice that $$f(x)>0 ,\forall x\in\Bbb{R}$$ $f(x)$ has no real root so $g(x)$ and $h(x)$ have no real root and are positive. Then from $g(a_j)h(a_j)=1$, we know $g(a_j)= h(a_j)=1$. So as $g(x)\neq1$,$$g(x)-1=(x-a_1)(x-a_2)\cdots(x-a_{14})$$Then $\deg(h)=14 $ and $h(x)$ should also be $(x-a_1)\cdots(x-a_{14})+1$, but this time $f(x)$ is not equal to $g(x)h(x)$, which yields a contradiction.