2012 AMC 10A Problem 22

The first $m$ odd integers add to $m^2$, and the first $n$ even integers add to $2(1 + 2 + \cdots + n) = n(n+1)$. The condition reads
$$m^2 = n^2 + n + 212.$$

The right side is nearly a perfect square. Multiply by $4$ so that $4n^2 + 4n + 1 = (2n+1)^2$ appears:
$$4m^2 = (2n+1)^2 + 847 \quad\Longrightarrow\quad (2m - 2n - 1)(2m + 2n + 1) = 847.$$

Factor $847 = 7 \cdot 121 = 7 \cdot 11^2$; its factor pairs are $1 \cdot 847$, $7 \cdot 121$ and $11 \cdot 77$. On the left, the second factor is the larger one, so for each pair $(S, L)$ we set $2m - 2n - 1 = S$ and $2m + 2n + 1 = L$. Subtracting gives $4n + 2 = L - S$, so $n = \frac{L - S - 2}{4}$ and $m = \frac{L + S}{4}$:

• $(1, 847)$: $n = 211$, $m = 212$.
• $(7, 121)$: $n = 28$, $m = 32$.
• $(11, 77)$: $n = 16$, $m = 22$.

All three are positive integers; for instance $22^2 - 16 \cdot 17 = 484 - 272 = 212$. The possible values of $n$ sum to $211 + 28 + 16 = 255$.

The answer is $\boxed{\textbf{(A)}\ 255}$.

Two quadratics in different variables with equal leading coefficients become a difference of squares once each is completed; scaling by $4$ avoids fractions. That converts an equation with infinitely many candidate pairs into a finite list of factor pairs of a single constant. Whenever an integer equation has the shape $A^2 - B^2 = \text{constant}$, factor and enumerate.

Put $d = m - n \ge 1$. Then $m^2 - n^2 = (m-n)(m+n) = d(2n + d)$ must equal $n + 212$, so $n(2d - 1) = 212 - d^2$ and $n = \frac{212 - d^2}{2d - 1}$. Testing $d = 1, 2, \dots, 14$ (beyond that the numerator is negative), only $d = 1, 4, 6$ give integers: $n = 211, 28, 16$. Same total, $255$.

• Using n^2 or n(n-1) for the sum of the first n even integers, or missing one of the three factor pairs of 847.
• Discarding the pair $1 \cdot 847$ as "too big" and answering $28 + 16 = 44$; $n = 211$, $m = 212$ is a legitimate solution.
• Assigning the larger factor to $2m - 2n - 1$, which produces negative $n$.