AMC 8 · 2014 · #13

Grade 4 number-theory

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

parityperfect-squareslogical-deduction caseworksystematic-enumeration ↑ Prerequisites: paritylogical-deduction

📏 Medium solution 💡 3 insights

Problem

n and m are integers whose squares add to an even number: n² + m² is even. Among the five statements about n and m, find the one that can never happen.

Pick an answer.

(A)

n and m are even

(B)

n and m are odd

(C)

n+m is even

(D)

n+m is odd

(E)

none of these are impossible

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

There are only four parity pairings for (n, m): (even, even), (even, odd), (odd, even), (odd, odd). Tool #2 (Make a Systematic List) writes them all down so none are missed. Tool #3 (Eliminate Possibilities) then uses the given condition n²+m² even to throw out the parity pairs that violate it, leaving only the survivors. Checking each answer choice against the survivors tells us which choice is impossible. We deliberately avoid heavier tools (algebra, formal proof) — a small parity table is the cleanest path.

1STEP 1

List all four even/odd combos for (n, m). Squaring never changes parity, so n² + m² matches the parity of n + m.

n & m & n²+m² ; even & even & even ; even & odd & odd ; odd & even & odd ; odd & odd & even

2STEP 2

Apply n² + m² even: drop the two odd-result rows. Survivors are (even, even) and (odd, odd), so n and m must share the same parity.

Survivors: (n,m)∈{(even,even),(odd,odd)}

3STEP 3

Test the survivors: (A) both even and (B) both odd each occur, and (C) n + m even holds in both — so A, B, C are all possible.

(A) ✓ (B) ✓ (C) ✓

4STEP 4

Now (D): an odd n + m needs one even and one odd, but survivors share parity — so n + m can never be odd here.

n+m odd → one even, one odd ∉ survivors → impossible

5STEP 5

(D) is impossible while (A), (B), (C) are possible, so (E) is false — the unique answer is (D).

(D) n+m is odd

Answer

n+m is odd

Sanity check with concrete numbers. (n,m)=(2,4): n²+m²=4+16=20 even, n+m=6 even — consistent with our rule. (n,m)=(1,3): 1+9=10 even, n+m=4 even — again consistent. (n,m)=(2,3): 4+9=13 odd — violates the given, so this pair is not allowed, and indeed n+m=5 is odd, exactly the case our argument rules out. Every example agrees: whenever n²+m² is even, n+m is forced to be even, so (D) is impossible.

💡Key takeaway

This AMC 8 problem only needs the Grade 4 even/odd rule — squaring keeps parity, so n²+m² and n+m are always the same parity!

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