AMC 10 · 2023 · #14

Grade 8 algebra

polynomial-factoringsymmetry-argumentbound-inequality-then-enumeratesystematic-enumeration convert-to-algebrabound-inequality-then-enumeratecasework ↑ Prerequisites: polynomial-factoringsystematic-enumeration

📏 Long solution 💡 3 insights

Problem

How many ordered pairs of integers $(m, n)$ satisfy the equation $m^2+mn+n^2 = m^2n^2$ ?

$\textbf{(A) }7\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }5$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Count the integer ordered pairs $(m, n)$ that satisfy $m^2 + mn + n^2 = m^2 n^2$.

Givens: Equation: $m^2 + mn + n^2 = m^2 n^2$; $m, n$ are integers (positive, negative, or zero); Ordered pairs: $(1, -1)$ and $(-1, 1)$ count as two distinct pairs; Answer choices: (A) $7$, (B) $1$, (C) $3$, (D) $6$, (E) $5$

Unknowns: Total number of integer ordered pairs $(m, n)$ satisfying the equation

Understand

Restated: Count the integer ordered pairs $(m, n)$ that satisfy $m^2 + mn + n^2 = m^2 n^2$.

Plan

Primary tool: #6 Guess and Check

Secondary: #2 Make a Systematic List, #13 Convert to Algebra, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities, #5 Look for a Pattern

Tool #6 (Guess and Check) gets early traction — try $(0,0), (1,1), (1,-1), (2,1), \ldots$ to see what works and develop a sense of how restrictive the equation is. Tool #2 (Systematic List) of small cases ($|m|, |n| \le 2$) finds every actual solution. Tool #13 (Algebra) is the closer: adding $mn$ to both sides gives $(m+n)^2 = mn(mn+1)$, i.e. two consecutive integers multiplying to a perfect square — only possible when one of them is zero. Tool #9 (Easier Problem) sanity-check: $m + n = m \cdot n$ has finitely many integer pairs; same flavor. Tool #3 (Eliminate): since small choices include $1$ and $3$, finding three solutions rules out $1$ and excludes large choices.

Execute — Answer: C

#6 Guess and Check 6.EE.A.2 Step 1

Guess-and-check a handful of small integer pairs to anchor intuition.
$(0, 0)$: $0 + 0 + 0 = 0 = 0 \cdot 0$.
Works.
$(1, 1)$: $1 + 1 + 1 = 3$, RHS $= 1$.
Fails.
$(1, -1)$: $1 - 1 + 1 = 1$, RHS $= 1$.
Works.
$(-1, 1)$: same value $1 = 1$.
Works.
$(2, 2)$: LHS $= 4 + 4 + 4 = 12$, RHS $= 16$.
Fails.
$(2, 1)$: LHS $= 4 + 2 + 1 = 7$, RHS $= 4$.
Fails.

$$(0, 0), (1, -1), (-1, 1) \text{ work; others tested fail.}$$

💡 Plugging in small integers either lands on a solution or shows how fast the RHS $m^2 n^2$ grows past the LHS — Grade 6 expression-evaluation.

#9 Solve an Easier Related Problem 8.EE.A.2 Step 2

Why does the LHS-versus-RHS gap explode for larger $|m|, |n|$?
For $|m|, |n| \ge 2$, the RHS $m^2 n^2 \ge 16$ but the LHS $m^2 + mn + n^2$ stays much smaller because $|mn| \le \tfrac{m^2 + n^2}{2}$ (AM-GM type bound), so LHS $\le \tfrac{3}{2}(m^2 + n^2)$.
Meanwhile RHS $= m^2 n^2 \ge 4(m^2 + n^2)$ when $|m|, |n| \ge 2$ and $m^2 + n^2 \ge 8$, so RHS $>$ LHS far away from the origin.
Solutions must be small.

$$|m|, |n| \ge 2 \Rightarrow \text{RHS} \gg \text{LHS}$$

💡 Quartic on the right outpaces quadratic on the left — solutions can only live near the origin, so a finite search suffices.

#13 Convert to Algebra 8.EE.A.2 Step 3

Make it rigorous with one algebraic trick.
Add $mn$ to both sides of the original equation to make the LHS a perfect square: $m^2 + mn + n^2 + mn = m^2 n^2 + mn$, i.e.
$(m + n)^2 = mn(mn + 1)$.

$$(m + n)^2 = mn \cdot (mn + 1)$$

💡 Adding $mn$ to both sides turns the LHS into $(m + n)^2$ — Grade 8 perfect-square trick.

#5 Look for a Pattern 8.EE.A.2 Step 4

Let $k = mn$.
The right side is $k(k+1)$ — the product of two consecutive integers.
The LHS $(m + n)^2$ is a perfect square.
So we need $k(k+1)$ to be a perfect square.
Two consecutive integers are coprime (their gcd divides their difference $1$), so for their product to be a perfect square, each factor must itself be a perfect square.
The only consecutive integers that are both perfect squares are $0$ and $1$ (because $j^2$ and $(j+1)^2$ differ by $2j + 1 > 1$ when $j \ge 1$, and only $0, 1$ are squares within a unit).
Hence $k(k+1) = 0$, i.e.
$k = 0$ or $k = -1$.

$$mn = 0 \text{ or } mn = -1$$

💡 Consecutive integers can only multiply to a perfect square when one of them is zero — a clean Grade 8 number-theory observation.

#2 Make a Systematic List 8.EE.C.7 Step 5

Solve each case for $(m, n)$ with $(m + n)^2 = 0$, i.e.
$m + n = 0$.
**Case A: $mn = 0$.** Then $m = 0$ or $n = 0$.
Combined with $m + n = 0$, this forces $m = n = 0$.
One solution: $(0, 0)$.
**Case B: $mn = -1$.** Then $m + n = 0$ gives $n = -m$, so $m \cdot (-m) = -1$, $m^2 = 1$, $m = \pm 1$.
Two ordered solutions: $(1, -1)$ and $(-1, 1)$.

$$\text{Case A: } (0, 0). \quad \text{Case B: } (1, -1), (-1, 1).$$

💡 Each case reduces to a one-variable linear or quadratic equation solvable by Grade 8 algebra.

#3 Eliminate Possibilities 8.EE.A.2 Step 6

Collect all solutions: $(0, 0), (1, -1), (-1, 1)$.
That is $3$ ordered pairs.

$$\#\{\text{pairs}\} = 3 \;\Rightarrow\; \textbf{(C)}$$

💡 Three ordered pairs matches choice (C).

[1] #6 6.EE.A.2 Guess-and-check a handful of small integer pairs to anchor intuition. $(0, 0)$:

[2] #9 8.EE.A.2 Why does the LHS-versus-RHS gap explode for larger $|m|, |n|$? For $|m|, |n| \ge

[3] #13 8.EE.A.2 Make it rigorous with one algebraic trick. Add $mn$ to both sides of the origina

[4] #5 8.EE.A.2 Let $k = mn$. The right side is $k(k+1)$ — the product of two consecutive intege

[5] #2 8.EE.C.7 Solve each case for $(m, n)$ with $(m + n)^2 = 0$, i.e. $m + n = 0$. **Case A: $

[6] #3 8.EE.A.2 Collect all solutions: $(0, 0), (1, -1), (-1, 1)$. That is $3$ ordered pairs.

Review

Reasonableness: Verify each solution by direct substitution. $(0, 0)$: LHS $= 0 + 0 + 0 = 0$, RHS $= 0 \cdot 0 = 0$. $(1, -1)$: LHS $= 1 + (-1) + 1 = 1$, RHS $= 1 \cdot 1 = 1$. $(-1, 1)$: same by symmetry. All three pass. Boundary sanity: did we miss $(0, k)$ for nonzero $k$? Then LHS $= 0 + 0 + k^2 = k^2 \ne 0 = 0 \cdot k^2 = \text{RHS}$ unless $k = 0$. So no other zero-row solutions. Did we miss $(m, n)$ with $|m|, |n| \ge 2$? The bound in Step 2 rules them out; the algebraic factoring in Steps 3-5 rules them out cleanly with no leftover cases.

Alternative: Tool #2 (Systematic List) over a finite box: from Step 2, all solutions have $|m|, |n| \le 2$. List the $25$ candidate pairs and check each — finds the same three. Slower but no algebra needed; works for a student who isn't yet comfortable with the SFFT-style manipulation in Step 3.

CCSS standards used (min grade 8)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Substituting small integer values of $(m, n)$ into the LHS and RHS to test candidate pairs.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Rewriting $m^2 + mn + n^2 + mn = (m + n)^2$, recognizing the perfect square, and reasoning about when a product of consecutive integers is a perfect square.)
8.EE.C.7 Solve linear equations in one variable (Solving $m + n = 0$ and $m^2 = 1$ to extract $(0, 0), (1, -1), (-1, 1)$.)

⭐ This AMC 10 problem only needs Grade 8 "completing the square" you already know — adding $mn$ to both sides turns it into $(m+n)^2 = mn(mn+1)$, and since two consecutive integers can only multiply to a perfect square when one of them is zero, the only solutions are $(0, 0), (1, -1), (-1, 1)$ — exactly $3$ pairs.