Track / Stage 3 / 248 of 260 #248 of 1964

Problem 248

AMC 10/12, early questions

Number theory Difficulty 3.9 Find the answer

Assume that $a$ , $b$ , $c$ , and $d$ are positive integers such that $a^5 = b^4$ , $c^3 = d^2$ , and $c - a = 19$ . Determine $d - b$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Official solution

It follows from the givens that $a$ is a perfect fourth power, $b$ is a perfect fifth power, $c$ is a perfect square and $d$ is a perfect cube. Thus, there exist integers $s$ and $t$ such that $a = t^4$ , $b = t^5$ , $c = s^2$ and $d = s^3$ . So $s^2 - t^4 = 19$ . We can factor the left-hand side of this equation as a difference of two squares, $(s - t^2)(s + t^2) = 19$ . 19 is a prime number and $s + t^2 > s - t^2$ so we must have $s + t^2 = 19$ and $s - t^2 = 1$ . Then $s = 10, t = 3$ and so $d = s^3 = 1000$ , $b = t^5 = 243$ and $d-b=\boxed{757}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.