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Problem 987

AMC 12 late, AIME early

Number theory Difficulty 4.8 Prove it Harvard-MIT Math Tournament · United States

If $x$ , $y$ , and $z$ are distinct positive integers such that $x^{2} + y^{2} = z^{3}$ , what is the smallest possible value of $x + y + z$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solutions — 2

Solution 1

Solution:
Without loss of generality let $x > y$ . We must have $z^{3}$ expressible as the sum of two squares, and this first happens when $z = 5$ . Then $x$ and $y$ can be $10$ and $5$ or $11$ and $2$ . If $z > 5$ then $z \geq 10$ for $z^{3}$ to be a sum of two distinct squares, so $x^{2} > 500$ , $x > 22$ , so $x + y + z > 32$ . Thus the smallest possible value of $x + y + z$ is $11 + 2 + 5 = \mathbf{18}$ .

Solution 2

Solution:
If $z > 5$ , then $z \geq 6$ , so $z^{3} \geq 216$ . Now $x^{2} + y^{2} \geq 216$ , so $x \geq 11$ and $y \geq 1$ , thus $x + y + z \geq 18$ . Since $x = 11$ , $y = 1$ , $z = 6$ does not work, we must have $x + y + z > 18$ , and the solution given is the best possible.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.