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I just thought of kind of a cool number theory/algebra problem.

Given that $$\sqrt{b^2-4ac}\in\Bbb N$$ To which sets must $a,b,c$ belong?

It is obvious that $$b^2-4ac\in\left\{x^2|x\in\Bbb N\right\}$$ But beyond that, I do not know what to do. May I have some help?

Edit:

To make things more interesting, what if $a,b,c\in\Bbb N$?

clathratus
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1 Answers1

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all triples $a,b,c$ that work are given by $$ ax^2 + b x y + c y^2 = (sx+ty)(ux+vy) \; , \; $$ $$ a = su, $$ $$ b = sv+tu, $$ $$ c = tv . $$ When those happen, $$ b^2 - 4ac = s^2 v^2 + 2 stuv + t^2 u^2 - 4 stuv =s^2 v^2 - 2 stuv + t^2 u^2 = (sv-tu)^2 $$

see Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable

Will Jagy
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