Question

Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17) = 10$ and $P(24) = 17.$ Given that the equation $P(n) = n + 3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1 \cdot n_2.$

Answer

Solution:

Let $S(x) = P(x) - x - 3,$ so $S(17) = S(24) = -10.$ Since $S(x) + 10$ has integer coefficients and vanishes at $17$ and $24,$ $S(x) = -10 + (x - 17)(x - 24)Q(x)$ for some polynomial $Q$ with integer coefficients.

If $P(n) = n + 3$ for an integer $n,$ then $(n-17)(n-24)Q(n) = 10,$ so $(n-17)(n-24)$ divides $10.$ The factors $n - 17$ and $n - 24$ are integers differing by $7$ whose product divides $10,$ so they are $\{2, -5\}$ or $\{5, -2\},$ giving $n = 19$ and $n = 22.$ Both occur, for example, for $P(x) = x - 7 - (x-17)(x-24).$

Hence $n_1 \cdot n_2 = 19 \cdot 22 = 418.$

2005 AIME II Problem 13

Solution:

Problem 13 in Other Years