AIME 2015 Problem 6

We call the three roots (some may be equal to one another) $x_{1}, x_{2}$ , and $x_{3}$ . Using Vieta's formulas, we get $x_1+x_2+x_3 = a$ ,

$x_1 \times x_2+x_1 \times x_3+x_2 \times x_3 = \frac{a^2-81}{2}$ , and

$x_1 \times x_2 \times x_3 = \frac{c}{2}$ .

Squaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \cdot x_1 \cdot x_2+2 \cdot x_1 \cdot x_3+2 \cdot x_2 \cdot x_3 = a^2$ .

We can then subtract twice our second equation to get $x_1^2+x_2^2+x_3^2 = a^2-2 \cdot \frac{a^2-81}{2}$ .

Simplifying the right side:

$a^2-2 \cdot \frac{a^2-81}{2}$

$a^2-a^2+81$

$81$

So, we know $x_1^2+x_2^2+x_3^2 = 81$ .

We can then list out all the triples of positive integers whose squares sum to $81$ :

We get $(1, 4, 8)$ , $(3, 6, 6)$ , and $(4, 4, 7)$ .

These triples give $a$ values of $13$ , $15$ , and $15$ , respectively, and $c$ values of $64$ , $216$ , and $224$ , respectively.

We know that Jon still found two possible values of $c$ when Steve told him the $a$ value, so the $a$ value must be $15$ . Thus, the two $c$ values are $216$ and $224$ , which sum to $\boxed{\text{440}}$ .