AMC 10 · 2020 · #8

Grade 6 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

perfect-squaresbound-inequality-then-enumerateparity identify-subproblemsbound-inequality-then-enumeratecasework ↑ Prerequisites: perfect-squaresexponents

📏 Medium solution 💡 2 insights

Problem

Find the number of ordered pairs of integers satisfying that x to the two thousand twentieth power plus y squared equals two y.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

infinitely many

How to solve

Strategy Identify Subproblems

Tool #7 reshapes the equation by completing the square: x²⁰²⁰ + (y-1)² = 1 — now both terms are nonnegative integers summing to 1. Tool #9 (easier) replaces the scary x²⁰²⁰ with the simpler observation "it's 0 when x = 0 and at least 1 otherwise (and exactly 1 only when x = ± 1)". Tool #2 then lists the two cases that achieve the sum 1. Tool #3 confirms against the choices.

1STEP 1

Complete the square

It becomes two squares summing to one.

x²⁰²⁰ + y² - 2y = 0 → x²⁰²⁰ + (y - 1)² = 1

2STEP 2

List the possible splits

The two values are zero and one.

{x²⁰²⁰, (y-1)²} = {0, 1}

3STEP 3

Solve the first split

The first split gives two pairs.

x = 0, y ∈ {0, 2} → (0, 0), (0, 2)

4STEP 4

Solve the second split

The even exponent allows both signs.

x ∈ {-1, 1}, y = 1 → (1, 1), (-1, 1)

5STEP 5

Add the solutions

Add them all.

2 + 2 = 4

6STEP 6

Match the choice

There are 4 ordered pairs.

4 → (D)

Answer

Verify each of the 4 pairs satisfies the original x²⁰²⁰ + y² = 2y: (0, 0): 0 + 0 = 0 = 2 · 0 ✓; (0, 2): 0 + 4 = 4 = 2 · 2 ✓; (1, 1): 1 + 1 = 2 = 2 · 1 ✓; (-1, 1): 1 + 1 = 2 = 2 · 1 ✓. Also confirm no others: any |x| ≥ 2 makes x²⁰²⁰ ≥ 2²⁰²⁰, vastly larger than 1, so (y-1)² would have to be negative — impossible. Choice (E) infinitely many is ruled out.

💡Key takeaway

This AMC 12 problem only needs Grade 6 expression rewriting you already know — complete the square to get x²⁰²⁰ + (y-1)² = 1, then notice two nonnegative whole numbers add to 1 only as 0 + 1 or 1 + 0. That gives (0, 0), (0, 2), (1, 1), (-1, 1) — 4 pairs.