AMC 8 · 2014 · #13
Grade 4 number-theoryProblem
If n and m are integers and n2+m2 is even, which of the following is impossible?
Pick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: $n$ and $m$ are integers whose squares add to an even number: $n^2 + m^2$ is even. Among the five statements about $n$ and $m$, find the one that can never happen.
Givens: $n$ and $m$ are integers (positive, negative, or zero); $n^2 + m^2$ is even; Answer choices: (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
Unknowns: Which of the five statements is impossible given the condition on $n^2+m^2$
Understand
Restated: $n$ and $m$ are integers whose squares add to an even number: $n^2 + m^2$ is even. Among the five statements about $n$ and $m$, find the one that can never happen.
Givens: $n$ and $m$ are integers (positive, negative, or zero); $n^2 + m^2$ is even; Answer choices: (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
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #2 Make a Systematic List
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 $n^2+m^2$ 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.
Execute — Answer: D
4.OA.B.4 Step 1 - List every parity combination for $(n, m)$ and the resulting parity of $n^2 + m^2$.
- Squaring does not change parity, so $n^2 + m^2$ has the same parity as the sum of an even/odd from $n$ and from $m$.
💡 Building the parity table from scratch is the Grade 4 "even and odd, factors and multiples" idea applied systematically.
3.OA.D.9 Step 2 - Use the given condition: $n^2 + m^2$ is even.
- Eliminate the two rows whose result is odd.
- Only the (even, even) and (odd, odd) rows survive — meaning $n$ and $m$ must have the same parity.
💡 Throwing out rows that contradict the given is the core eliminate-possibilities move, and the pattern "squares keep parity, so sum-of-squares parity matches $n+m$ parity" is Grade 3 pattern reasoning.
4.OA.B.4 Step 3 - Check each answer choice against the two surviving cases.
- (A) both even — happens in row 1, possible.
- (B) both odd — happens in row 4, possible.
- (C) $n+m$ even — even$+$even and odd$+$odd are both even, so it holds in both survivors, possible.
💡 Verifying each choice against the survivor list is a clean elimination check using even/odd rules.
4.OA.B.4 Step 4 - Now test (D) $n+m$ is odd.
- An odd sum requires one even and one odd integer, but the survivors force $n$ and $m$ to share parity.
- So $n+m$ can never be odd under the given condition.
💡 Same-parity pairs always sum to an even number, so an odd sum is ruled out — the eliminate-possibilities tool delivers the answer directly.
4.OA.B.4 Step 5 - Since (D) is impossible and (A), (B), (C) are all possible, (E) "none impossible" is false.
- The unique answer is (D).
💡 Exactly one choice survives the elimination — that is the impossible scenario the problem asks for.
4.OA.B.4 List every parity combination for $(n, m)$ and the resulting parity of $n^2 + m^ 3.OA.D.9 Use the given condition: $n^2 + m^2$ is even. Eliminate the two rows whose resul 4.OA.B.4 Check each answer choice against the two surviving cases. (A) both even — happen 4.OA.B.4 Now test (D) $n+m$ is odd. An odd sum requires one even and one odd integer, but 4.OA.B.4 Since (D) is impossible and (A), (B), (C) are all possible, (E) "none impossible Review
Reasonableness: Sanity check with concrete numbers. $(n,m)=(2,4)$: $n^2+m^2=4+16=20$ 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 $n^2+m^2$ is even, $n+m$ is forced to be even, so (D) is impossible.
Alternative: Tool #5 (Look for a Pattern) using a key shortcut: for any integer $k$, $k^2$ and $k$ have the same parity, so $n^2+m^2$ and $n+m$ also share parity. The given says $n^2+m^2$ is even, which forces $n+m$ to be even. Therefore $n+m$ odd is impossible — choice (D) — with no casework needed.
CCSS standards used (min grade 4)
3.OA.D.9Identify arithmetic patterns (including patterns in the addition table), and explain them using properties of operations (Recognizing the even/odd pattern that even+even and odd+odd give an even sum, while even+odd gives an odd sum — the parity rule that drives the elimination.)4.OA.B.4Find all factor pairs and recognize multiples; determine whether a given whole number is a multiple of a given one-digit number (including even/odd) (Classifying each integer as even or odd, using that a square has the same parity as the integer, and checking each answer choice against the surviving (same-parity) cases.)
⭐ This AMC 8 problem only needs the Grade 4 even/odd rule — squaring keeps parity, so $n^2+m^2$ and $n+m$ are always the same parity!
⭐ This AMC 8 problem only needs the Grade 4 even/odd rule — squaring keeps parity, so $n^2+m^2$ and $n+m$ are always the same parity!
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