Is it true that if for some quadratic $ax^2 + bx + c$ (where $a, b, c$ are integers) there exists 2 integers $m$ and $n$ such that $m + n = b$ and $mn = ac$, then there exists integers $A, B, C, D$ such that $(Ax + B)(Cx + D) = ax^2 + bx + c$?
I'm able to prove the converse that if that $ax^2 + bx + c = (Ax + B)(Cx + D)$ (where $a, b, c$ are real and $A, B, C, D$ are integers), then there exists there exists 2 integers $m$ and $n$ such that $m + n = b$ and $mn = ac$, namely where $m = AD$ and $n = BC$. This is because $a = AC$, $b = AD + BC$, and $c = BD$ (which also implies $a, b, c$ are integers), meaning $m + n = AD + BC = b$ and $mn = (AD)(BC) = (AC)(BD) = ac$ .
However, when I try to prove the previous claim, I get that I can factor the quadratic as either $(mx + c)(\frac{n}{c}x + 1)$ or $(nx + c)(\frac{m}{c}x + 1)$, however how can I show either n/c or m/c is an integer? Do we need some use of number theory or clever algebra that I'm not seeing here or something?
I'm just trying to better understand basic factoring techniques I learned in grade school but never really thought why do they work or make sense. Since with the claim I was able to prove, it's contrapositive tells me if you can't find 2 integers $m$ and $n$ such that $m + n = b$ and $mn = ac$, then there doesn't not exist 4 integers $A, B, C, D$ such that $(Ax + B)(Cx + D) = ax^2 + bx + c$ (correct?)
But then this doesn't prove to me the claim I ask about since an implication's converse is not necessarily true. Hence I kindly need some help here.
By the way this was inspired by Khan Academy's "proof" (don't think it actually proves my claim I'm asking about) that I ironically never learned about in middle school when 1st learning this technique (maybe I wasn't paying attention, I don't know): https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-quadratics-grouping/a/factoring-quadratics-leading-coefficient-not-1!
