Question

For integers $a,$ $b,$ $c,$ and $d,$ let $f(x) = x^2 + ax + b$ and $g(x) = x^2 + cx + d.$ Find the number of ordered triples $(a, b, c)$ of integers with absolute values not exceeding $10$ for which there is an integer $d$ such that $g(f(2)) = g(f(4)) = 0.$

Answer

Solution:

The condition says the integers $f(2) = 4 + 2a + b$ and $f(4) = 16 + 4a + b$ are both roots of the monic quadratic $g.$ These two values are equal exactly when $a = -6.$

If $a = -6,$ then for any $b$ and any $c$ the choice $d = -f(2)^2 - c\,f(2)$ makes $f(2) = f(4)$ a root of $g,$ giving $21 \cdot 21 = 441$ triples. If $a \ne -6,$ the two distinct values must be the two roots of $g,$ so Vieta forces $c = -(f(2) + f(4)) = -(20 + 6a + 2b),$ and then $d = f(2)f(4)$ is an integer. The requirement $|c| \le 10$ becomes $-15 \le 3a + b \le -5.$

For each $a,$ count integers $b \in [-10, 10]$ with $-15 - 3a \le b \le -5 - 3a:$ the counts are $2, 5, 11, 11, 11, 11, 9, 6, 3$ for $a = -8, -7, -5, -4, -3, -2, -1, 0, 1$ respectively, and $0$ for all other $a \ne -6,$ totaling $69.$ The answer is $441 + 69 = 510.$

2020 AIME I Problem 11

Solution:

Problem 11 in Other Years