AMC 10 · 2023 · #14

Grade 8 algebra

polynomial-factoringsymmetry-argumentbound-inequality-then-enumeratesystematic-enumeration convert-to-algebrabound-inequality-then-enumeratecasework ↑ Prerequisites: polynomial-factoringsystematic-enumeration

📏 Long solution 💡 3 insights

Problem

Count the integer ordered pairs (m, n) that satisfy m² + mn + n² = m² n².

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

Tool #6 (Guess and Check) gets early traction — try (0,0), (1,1), (1,-1), (2,1), … to see what works and develop a sense of how restrictive the equation is. Tool #2 (Systematic List) of small cases (|m|, |n| ≤ 2) finds every actual solution. Tool #13 (Algebra) is the closer: adding mn to both sides gives (m+n)² = mn(mn+1), i.e. two consecutive integers multiplying to a perfect square — only possible when one of them is zero. Tool #9 (Easier Problem) sanity-check: m + n = m · n has finitely many integer pairs; same flavor. Tool #3 (Eliminate): since small choices include 1 and 3, finding three solutions rules out 1 and excludes large choices.

1STEP 1

Try small pairs: (0, 0), (1, -1), (-1, 1) satisfy it, while (1, 1), (2, 2), (2, 1) all fail.

(0, 0), (1, -1), (-1, 1) work; others tested fail.

2STEP 2

For |m|, |n| ≥ 2 the right side (mn)² outgrows the left, so any solution must sit near the origin — a finite search suffices.

|m|, |n| ≥ 2 → RHS ≫ LHS

3STEP 3

Add mn to both sides so the left becomes a perfect square, giving (m + n)² = mn(mn + 1).

(m + n)² = mn · (mn + 1)

4STEP 4

mn and mn + 1 are consecutive, so their product is a perfect square only when it is zero — forcing mn = 0 or mn = -1.

mn = 0 or mn = -1

5STEP 5

Both cases give m + n = 0: mn = 0 forces (0, 0); mn = -1 gives m² = 1, so (1, -1) and (-1, 1).

Case A: (0, 0). Case B: (1, -1), (-1, 1).

6STEP 6

The complete list is (0, 0), (1, -1), (-1, 1) — three ordered pairs, choice (C).

#{pairs} = 3 → (C)

Answer

Verify each solution by direct substitution. (0, 0): LHS = 0 + 0 + 0 = 0, RHS = 0 · 0 = 0. (1, -1): LHS = 1 + (-1) + 1 = 1, RHS = 1 · 1 = 1. (-1, 1): same by symmetry. All three pass. Boundary sanity: did we miss (0, k) for nonzero k? Then LHS = 0 + 0 + k² = k² ≠ 0 = 0 · k² = RHS unless k = 0. So no other zero-row solutions. Did we miss (m, n) with |m|, |n| ≥ 2? The bound in Step 2 rules them out; the algebraic factoring in Steps 3-5 rules them out cleanly with no leftover cases.

💡Key takeaway

This AMC 10 problem only needs Grade 8 "completing the square" you already know — adding mn to both sides turns it into (m+n)² = mn(mn+1), and since two consecutive integers can only multiply to a perfect square when one of them is zero, the only solutions are (0, 0), (1, -1), (-1, 1) — exactly 3 pairs.