Dropped AIME Problem
Consider $x,y \in \mathbb{R}$ such that $$ |x| \neq |y|, \ x^{3}=15x+4y, \text{ and } y^{3}=4x+15y.$$ Find $x^{2}+y^{2}.$

Solution: Well, observe that $$x^{3} + y^{3} = 19x+19y=19(x+y)=(x+y)(x^{2}-xy+y^{2}).$$ Thus, since $|x| \neq |y|,$ $$x^{2}+y^{2}=19+xy.$$ We also observe that $$x^{3}-y^{3}=11x-11y=11(x-y)=(x-y)(x^{2}+xy+y^{2}).$$ Thus, since $|x|\neq|y|,$ we have that $$x^{2}+y^{2}=11-xy.$$ Thus, $$19+xy=11-xy \implies 2xy=-8 \implies xy = -4.$$ Thus, $x^{2}+y^{2} = 15.$

Demystifying the AC Method
Let’s explore the AC method for factoring quadratic expressions, particularly when the leading coefficient $a$ is unequal to $1$. We consider the anatomy of $$ax^{2}+bx+c,$$ where $a \neq 0$. If this expression is factorable, then it can be written as a product of two linear factors, like so: $$ax^{2}+bx+c = (mx+p)(nx+q).$$ Expanding the right-hand side gives us $$ax^{2}+bx+c = mnx^{2} + (mq+np)x+pq.$$ Then, it must be the case that $a=mn$, $b=mq + np,$ and $c=pq.$ We recall that the AC method states that one would have to multiply $a$ and $c$, then find factors of $ac$ such that their sum is equal to $b$. From above, we see that $ac=mnpq=mqnp=(mq)(np)$, and so $b$ is indeed the sum of two factors of $ac$.