Track / Stage 4 / 302 of 340 #562 of 1964

Problem 562

AMC 12 late, AIME early

Number theory Difficulty 5.0 Find the answer

Example 1 Let $N=23x+92y$ be a perfect square, and $N$ does not exceed 2392. Then the number of all positive integer pairs $(x, y)$ that satisfy the above conditions is $\qquad$ pairs.

Your answer

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

Official solution

Solution: Since $N=23(x+4 y)$ , and 23 is a prime number, there exists a positive integer $k$ , such that $x+4 y=23 k^{2}$ .
Also, because $23(x+4 y)=N \leqslant 2392$ , we have $x+4 y \leqslant 104$ .
Therefore, $23 k^{2}=x+4 y \leqslant 104$ , which means $k^{2} \leqslant 4$ .
Thus, $k^{2}=1,4$ .
When $k^{2}=1$ , $x+4 y=23$ , in this case, $y \leqslant 5$ , we can get 5 solutions;

When $k^{2}=4$ , $x+4 y=92$ , in this case, $y \leqslant 22$ , we can get 22 solutions.

In summary, there are a total of $5+22=27$ pairs of positive integers $(x, y)$ that satisfy the conditions.

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