🇺🇸 AMC 8 ⇄ switch contest
2014 AMC 8

Problem 13

Problem 13 · 2014 AMC 8 Medium
Number Theory parity

If n and m are integers and n2 + m2 is even, which of the following is impossible?

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Answer: D — n + m cannot be odd.
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Hint 1 of 2
Squaring never changes even/odd: an even number stays even when squared, an odd stays odd. So n2+m2 behaves exactly like n+m for parity — the exponents are a red herring.
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Hint 2 of 2
An even sum forces n and m to share parity. Now check which listed statement that rules out.
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Approach: a square keeps the parity of its base
  1. Squaring preserves parity, so n2+m2 even means n+m is even too ⇒ n and m have the same parity (both even or both odd).
  2. Same parity always sums to an even number, so n + m can never be odd — that's the impossible choice.
  3. Sanity check on the others: both even (2,2) works; both odd (1,1) works; an even sum is exactly what's forced. Only "sum is odd" is impossible.
  4. Takeaway: for parity questions you can erase squares, products, and other parity-preserving steps and just track even/odd directly.
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