AMC 10 · 2024 · #11
Grade 8 algebranumber-theoryProblem
How many ordered pairs of integers (m,n) satisfy n2−49=m?
Pick an answer.
AMC 10 2024 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: Count the ordered pairs of integers $(m, n)$ that satisfy $\sqrt{n^2 - 49} = m$.
Givens: The equation $\sqrt{n^2 - 49} = m$; Both $m$ and $n$ are integers; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) infinitely many
Unknowns: The number of ordered integer pairs $(m, n)$ that satisfy the equation
Understand
Restated: Count the ordered pairs of integers $(m, n)$ that satisfy $\sqrt{n^2 - 49} = m$.
Givens: The equation $\sqrt{n^2 - 49} = m$; Both $m$ and $n$ are integers; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) infinitely many
Plan
Primary tool: #2 Make an Organized List
Secondary: #6 Guess and Check, #16 Change Focus / Count the Complement
Squaring the equation turns $\sqrt{n^2 - 49} = m$ into $n^2 - m^2 = 49$. Tool #2 (Make an Organized List) is the natural way to hunt for integer solutions: walk through small candidate values of $n$ with $|n| \ge 7$ and check whether $n^2 - 49$ is a perfect square. Tool #6 (Guess and Check) is what each test inside the list actually does. Tool #16 (Count the Complement) helps us avoid double work: because the equation only mentions $n^2$ and $m^2$, every solution with positive $n$ has a twin with negative $n$, so we can count positive $n$ first and double when appropriate.
Execute — Answer: D
8.EE.A.2 Step 1 - Square both sides to remove the radical.
- Because $\sqrt{\cdot}$ produces a non-negative result and equals the integer $m$, we know $m \ge 0$.
- Squaring is safe here.
💡 Grade 8 treats the square root symbol as the inverse of squaring a non-negative number — squaring both sides is the standard move to clear it.
6.EE.B.5 Step 2 - Reframe the search.
- The equation $n^2 - 49 = m^2$ says: find integers $n$ with $n^2 - 49$ equal to a perfect square.
- The square root forces $|n| \ge 7$, so start the list at $n = 7$ and march upward.
💡 Grade 6 framing of "solve an equation" as testing values that make both sides equal — here we test which $n$ make $n^2 - 49$ land on a perfect square.
4.OA.B.4 Step 3 - Walk up the positive values of $n$ and test each.
- Stop the search as soon as the gap $n^2 - (n-1)^2 = 2n - 1$ exceeds $49$, because once consecutive squares are more than $49$ apart, $n^2 - 49$ falls strictly between $(n-1)^2$ and $n^2$ and can never be a perfect square again.
- That bound is $2n - 1 > 49$, i.e.
- $n \ge 26$, so we only need to check $n = 7, 8, 9, \ldots, 25$.
💡 Grade 4 recognizing perfect squares ($0, 1, 4, 9, \ldots, 576$) is enough to scan the list. The gap argument prunes the search to a finite range.
6.NS.C.6 Step 4 - Mirror to negative $n$.
- The equation only sees $n^2$, so if $n$ works then $-n$ works with the same $m$.
- Positive $n$ gave two solutions ($n=7$ and $n=25$), so altogether we get four ordered pairs.
💡 Grade 6 idea that a number and its opposite have the same square — so every positive-$n$ solution is paired with a negative-$n$ twin.
8.EE.A.2 Square both sides to remove the radical. Because $\sqrt{\cdot}$ produces a non-n 6.EE.B.5 Reframe the search. The equation $n^2 - 49 = m^2$ says: find integers $n$ with $ 4.OA.B.4 Walk up the positive values of $n$ and test each. Stop the search as soon as the 6.NS.C.6 Mirror to negative $n$. The equation only sees $n^2$, so if $n$ works then $-n$ Review
Reasonableness: Verify each pair in the original equation. $(m, n) = (0, 7)$: $\sqrt{49 - 49} = \sqrt{0} = 0 = m$. Check. $(m, n) = (0, -7)$: $\sqrt{(-7)^2 - 49} = \sqrt{0} = 0 = m$. Check. $(m, n) = (24, 25)$: $\sqrt{625 - 49} = \sqrt{576} = 24 = m$. Check. $(m, n) = (24, -25)$: $\sqrt{625 - 49} = 24 = m$. Check. All four pairs satisfy the equation, and the search bound $n < 26$ rules out any larger $n$, so 4 is exact — answer (D).
Alternative: Tool #5 (Look for a Pattern) plus difference of squares: rewrite $n^2 - m^2 = 49$ as $(n-m)(n+m) = 49$. Since $49 = 1 \times 49 = 7 \times 7$ (and the same with both factors negated), and $m \ge 0$ forces $n + m \ge n - m$, the only factor pairs that produce integer $(m, n)$ are $(1, 49)$, $(7, 7)$, $(-49, -1)$, $(-7, -7)$. Solving each pair gives $(24, 25), (0, 7), (24, -25), (0, -7)$ — the same four pairs.
CCSS standards used (min grade 8)
8.EE.A.2Use square root and cube root symbols to represent solutions and evaluate square roots of small perfect squares (Reading $\sqrt{n^2 - 49} = m$ as "$m$ is the non-negative number whose square is $n^2 - 49$", which justifies squaring both sides to get $n^2 - 49 = m^2$ and forces $m \ge 0$.)6.EE.B.5Understand solving an equation as a process of answering which values make the equation true (Reframing the question as "which integers $n$ with $|n| \ge 7$ make $n^2 - 49$ a perfect square?" so the search can be done by testing values.)4.OA.B.4Find all factor pairs and recognize multiples; identify perfect squares within the range (Scanning $n = 7, 8, \ldots, 25$ and recognizing which values of $n^2 - 49$ ($0$ and $576$) are perfect squares.)6.NS.C.6Understand a rational number as a point on the number line and recognize opposite signs (Pairing each positive-$n$ solution with its negative-$n$ twin, since $n^2 = (-n)^2$.)
⭐ When a square root has to equal an integer, square both sides and hunt: which integers make the inside a perfect square? Bounding the search with the gap between consecutive squares turns an infinite-looking problem into a quick finite list.
⭐ When a square root has to equal an integer, square both sides and hunt: which integers make the inside a perfect square? Bounding the search with the gap between consecutive squares turns an infinite-looking problem into a quick finite list.
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