From E.J. Barbeau's Polynomials, the question states:
Let a,b,c be nonzero integers such that $(b,ac)=1$. Prove that $ax^2+bx±c$ can both be written as the product of linear polynomials with integer coefficients if and only if $ac=rs(r^2−s^2)$ and $b^2=(r^2+s^2)^2$, where $r$ and $s$ are coprime.
My approach was to first plug in the values and check the zeroes, giving me $$x = \frac{\pm(r^2 + s^2) \pm (s^2 - r^2 + 2rs)}{2a}$$ for $ax^2 + bx + c$. But I'm not really sure as to how I can proceed from here, but I guess it has to do with something using the divisibility conditions regarding $a$ stated in the question because using this as is gives me a constant along with the two linear polynomials when I factorize. I also don't know how to go about proving this in the other direction of the iff. A hint would be greatly appreciated.
COMMENT.-I have pay attention to this problem and almost proved it I remark that there are trinomes $ax^2+bx+c$ which are the product of linear polynomials of $\mathbb Z[x]$ but for which $b$ cannot be written as the sum of two squares. In fact, for example,$$15x^2+19x+6=(3x+2)(5x+3)$$ According to the proposed problem here we must have $19^2=(r^2+s^2)^2$ with $(r,s)=1$ so $19=r^2+s^2$ but $19$ is a prime number congruent to $-1$ modulo $4$ so it is known that cannot be written as a sum of two squares (easy to verify anyway).
It is true that $19^2=19^2+0^2$ on the context of the proof of an integer $n$ as a sum of two squares but assuming this in the given example we would have $ac=0$.
Nevertheless the problem could be valid if certain change on data is made.