Let's assume that $P$ is reducible in $\mathbb{Z}[X].$ There exist $F, G \in \mathbb{Z}[X], \text{deg}(F) \geq 1$ and $\text{deg}(G) \geq 1,$ such that \[P = (X-a_{1})^{2}(X-a_{2})^{2}\cdots (X-a_{n})^{2}+1 = F(X)G(X).\] We may take $F$ and $G$ to be monic. Let $k = \text{deg}(F), l = \text{deg}(G),$ so that $k+l=2n.$ Clearly, for any real $x, P(x) \geq 1.$ Hence $F$ and $G$ take positive values on $\mathbb{R}.$ For any $i, 1 \leq i \leq n, F(a_{i})G(a_{i}) = 1 \ \ (*).$ Therefore, for any $i, F(a_{i}) = G(a_{i}) = 1.$ If $k<l,$ then $k<n.$ But the polynomial $F-1$ is the zero polynomial. Hence $F=1,$ contradiction. Similarly $l<k$ is impossible. Thus $k=l=n.$ Thanks to $(*),$ for any $i, 1 \leq i \leq n, F(a_{i}) = G(a_{i}).$ So $F-G = 0,$ that is to say $F=G,$ so that $P(X) = F(X)^{2},$ which means that \[[F-(X-a_{1}) \cdots (X-a_{n})][F+(X-a_{1}) \cdots (X-a_{n})] = 1,\] which is impossible. Therefore $P$ is irreducible in $\mathbb{Z}[X].$ Remark : Actually, this polynomial is irreducible in $\mathbb{Q}[X],$ thanks to Gauss' lemma.

mathmanman wrote: Hence $F$ and $G$ take positive values on $\mathbb{R}.$ Why is this true? Isn't it possible that $F(X)$ and $G(X)$ are both negative for some $X\in{R}$?

They are both monic polynomials without real roots, thus they are everywhere positive (otherwise the intermediate value theorem would give some root).