2024 AMC 10A Problem 11

1. What do we know about $m$? It equals a square root, and a square root is never negative. So $m \ge 0$.
2. Get rid of the square root. Square both sides: $n^2 - 49 = m^2$, so
$$n^2 - m^2 = 49.$$
3. Factor. A difference of two squares factors: $n^2 - m^2 = (n - m)(n + m)$. So
$$(n - m)(n + m) = 49.$$
4. List the ways to write $49$ as a product of two integers. Since $m \ge 0$, the second factor $n + m$ is at least as large as the first factor $n - m$. The possibilities are:

($n$ is halfway between the two factors, and $m$ is half the gap between them.) The other orders, such as $n - m = 49$ and $n + m = 1$, would give a negative $m$, which is not allowed.
5. Check. $\sqrt{25^2 - 49} = \sqrt{625 - 49} = \sqrt{576} = 24$, and $\sqrt{7^2 - 49} = \sqrt{0} = 0$. The negative values of $n$ give the same squares.

So there are $4$ pairs: $(m, n) = (24, 25), (24, -25), (0, 7), (0, -7)$.

The answer is $\boxed{\textbf{(D)}\ 4}$.

"Square minus square equals a number" should make you factor it as (difference)(sum). Then finding whole-number answers is the same as listing the ways to split the number into two factors. The two details the answer choices test are that $n$ may be negative and $m$ may not.

Because $m$ is a square root, $m \ge 0$. Squaring, $n^2 - 49 = m^2$, so
$$(n - m)(n + m) = 49.$$
The two factors are integers whose product is $49$: the ordered factor pairs are $(1, 49), (7, 7), (49, 1), (-1, -49), (-7, -7), (-49, -1)$. Solving $n - m = d$, $n + m = 49/d$ for each:

• $(1, 49)$: $n = 25, m = 24$. Valid.
• $(7, 7)$: $n = 7, m = 0$. Valid.
• $(49, 1)$: $n = 25, m = -24$. Rejected ($m < 0$).
• $(-1, -49)$: $n = -25, m = -24$. Rejected.
• $(-7, -7)$: $n = -7, m = 0$. Valid.
• $(-49, -1)$: $n = -25, m = 24$. Valid.

Check: $\sqrt{625 - 49} = \sqrt{576} = 24$ and $\sqrt{49 - 49} = 0$. Four ordered pairs: $(24, 25), (24, -25), (0, 7), (0, -7)$.

The answer is $\boxed{\textbf{(D)}\ 4}$.