HMMT February

Algebra Difficulty 6.0 AIME, harder Find the answer

For each prime $p$ , a polynomial $P(x)$ with rational coefficients is called $p$ -good if and only if there exist three integers $a, b$ , and $c$ such that $0 \leq a<b<c<\frac{p}{3}$ and $p$ divides all the numerators of $P(a)$ , $P(b)$ , and $P(c)$ , when written in simplest form. Compute the number of ordered pairs $(r, s)$ of rational numbers such that the polynomial $x^{3}+10x^{2}+rx+s$ is $p$ -good for infinitely many primes $p$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

By Vieta, the sum of the roots is $-10(\bmod p)$ . However, since the three roots are less than $p/3$ , it follows that the roots are $\left(p-a^{\prime}\right)/3,\left(p-b^{\prime}\right)/3,\left(p-c^{\prime}\right)/3$ , where there are finitely many choices $a^{\prime}<b^{\prime}<c^{\prime}$ . By pigeonhole, one choice, say $(u, v, w)$ must occur for infinitely many $p$ . We then get that the roots of $P$ are $-u/3,-v/3$ , and $-w/3$ . Moreover, we must have that $u, v, w$ are all $1(\bmod 3)$ or all $2(\bmod 3)$ , and by Vieta, we have $u+v+w=30$ . The polynomial is then uniquely determined by $u, v, w$ . Thus, it suffices to count triples $u<v<w$ of positive integers such that $u, v, w$ are all $1(\bmod 3)$ or all $2(\bmod 3)$ and that $u+v+w=30$ . It's not very hard to list them all now. When $u, v, w \equiv 1(\bmod 3)$ , there are 7 triples: $(1,4,25),(1,7,22),(1,10,19),(1,13,16),(4,7,19)$ , $(4,10,16)$ , and $(7,10,13)$ . When $u, v, w \equiv 2(\bmod 3)$ , there are 5 triples: $(2,5,23),(2,8,20),(2,11,17),(5,8,17)$ , and $(5,11,14)$ . Hence, the answer is $7+5=12$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.