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1985 AIME Problems/Problem 7

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Problem

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 $d - b$ .

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 $d - b = 757$ .

See also

1985 AIME (Problems • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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