2024 AMC 10A Problem 11

Attempt Problem 11 of the 2024 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

How many ordered pairs of integers

(m, n)

satisfy

\sqrt{n^2 - 49} = m?

1

2

3

4

Infinitely many

Answer: D

Concepts:Diophantine Equation difference of squares radical

Difficulty rating: 1440

Small hint:

Squaring gives

n^2 - 49 = m^2,

so

(n - m)(n + m) = 49

Big hint:

49

has few factorizations; also

m = \sqrt{\cdots} \ge 0,

while

n

may be negative

Solution:

Note

m = \sqrt{n^2 - 49} \ge 0

has to be an integer, so

n^2 - 49 = m^2,

which means

(n - m)(n + m) = 49.

The factorizations of

49

give

|n| = 25, m = 24

or

|n| = 7, m = 0.

So the ordered pairs

(m, n)

are

(24, 25),

(24, -25),

(0, 7),

(0, -7).

That’s

4

of them. Thus, D is the correct answer.