By the Quadratic Formula, the zeros of $f(X)$ (where $f(X) \in \mathbb{Z}[X]$ is the field of polynomials with integer coefficients) are: $X = \frac{-a \pm \sqrt{a^2-4b}}{2}$ (i). The zeros are integral $\Leftrightarrow a^2-4b = N^2 \Rightarrow a^2 = N^2 + 4b$ (ii). Applying Pythagorean triplets to (ii) such that $N^2 = (p^2-q^2)^2, 4b = (2pq)^2, a^2 = (p^2+q^2)^2$ (for integers $p > q$) will produce: $X = \frac{-(p^2+q^2) \pm (p^2-q^2)}{2} \Rightarrow X = -p^2, -q^2$ (iii); or $f(X) = (X+p^2)(X+q^2) = X^2 + (p^2+q^2)X + p^2q^2$. If $n | f(X)$ for all $n \in \mathbb{N}$, then at least one of $p$ or $q$ is divisible by $n$. This gives us two cases to consider: CASE I ($p = \alpha n$ for some $\alpha \in \mathbb{Z}$): This produces $\frac{X^2 +(\alpha^{2}n^2+q^2)X + \alpha^{2}n^2q^2}{n} = \alpha^{2}n(X+q^2) + \frac{X^2 + q^2X}{n}$, which is divisible by $n \Leftrightarrow \exists u_{n} = kn$ for $k \in \mathbb{Z}$ and $n|f(u_{n})$. CASE II ($p =\alpha n, q = \beta n$ for some $\alpha, \beta \in \mathbb{Z}$): This produces $\frac{X^2 +(\alpha^{2}+\beta^{2})n^{2}X + \alpha^{2}\beta^{2}n^2}{n} = [\alpha^{2}\beta^{2} + (\alpha^{2}+\beta^{2})X]n + \frac{X^2}{n}$, which is divisible by $n \Leftrightarrow \exists u_{n} = kn$ for $k \in \mathbb{Z}$ and $n|f(u_{n})$. QED