2023 AMC 10B Problem 14

Attempt Problem 14 of the 2023 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 10B solutions, or check the answer key.

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14.

How many ordered pairs of integers (m,n)(m, n) satisfy the equation m2+mn+n2=m2n2?m^2 + mn + n^2 = m^2 n^2?

77

11

33

66

55

Answer: C
Concepts:Diophantine Equationbounding to limit casescasework
Difficulty rating: 1660
Solution:

If m=0,m = 0, the equation forces n2=0,n^2 = 0, giving (0,0).(0, 0). Otherwise both are nonzero; assume mn.|m| \le |n|. Then m2n2=m2+mn+n23n2,m^2 n^2 = m^2 + mn + n^2 \le 3n^2, so m23m^2 \le 3 and m=±1.m = \pm 1. Take m=1:m = 1: 1+n+n2=n21 + n + n^2 = n^2 gives n=1.n = -1. Take m=1:m = -1: n=1.n = 1. That leaves (0,0),(1,1),(1,1),(0,0), (1,-1), (-1,1), three in all. Therefore, the answer is C.

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