AMC 10 Step by Step

2024 AMC 10A Problem 11

P11-15~3 minPrint

How many ordered pairs of integers (m,n)(m, n) satisfy n2−49=m\sqrt{n^2 - 49} = m ?

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Key insight

Square both sides and factor: (n − m)(n + m) = 49. Then list the ways to split 49 into two integer factors.

Solution

1. What do we know about mm? It equals a square root, and a square root is never negative. So m≥0m \ge 0.
2. Get rid of the square root. Square both sides: n2−49=m2n^2 - 49 = m^2, so
n2−m2=49.n^2 - m^2 = 49.
3. Factor. A difference of two squares factors: n2−m2=(n−m)(n+m)n^2 - m^2 = (n - m)(n + m). So
(n−m)(n+m)=49.(n - m)(n + m) = 49.
4. List the ways to write 4949 as a product of two integers. Since m≥0m \ge 0, the second factor n+mn + m is at least as large as the first factor n−mn - m. The possibilities are:

n−mn - mn+mn + mnn (average of the two)mm
11494925252424
77777700
−49-49−1-1−25-252424
−7-7−7-7−7-700

(nn is halfway between the two factors, and mm is half the gap between them.) The other orders, such as n−m=49n - m = 49 and n+m=1n + m = 1, would give a negative mm, which is not allowed.
5. Check. 252−49=625−49=576=24\sqrt{25^2 - 49} = \sqrt{625 - 49} = \sqrt{576} = 24, and 72−49=0=0\sqrt{7^2 - 49} = \sqrt{0} = 0. The negative values of nn give the same squares.

So there are 44 pairs: (m,n)=(24,25),(24,−25),(0,7),(0,−7)(m, n) = (24, 25), (24, -25), (0, 7), (0, -7).

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

Why this works

"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 nn may be negative and mm may not.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Because mm is a square root, m≥0m \ge 0. Squaring, n2−49=m2n^2 - 49 = m^2, so
(n−m)(n+m)=49.(n - m)(n + m) = 49.
The two factors are integers whose product is 4949: the ordered factor pairs are (1,49),(7,7),(49,1),(−1,−49),(−7,−7),(−49,−1)(1, 49), (7, 7), (49, 1), (-1, -49), (-7, -7), (-49, -1). Solving n−m=dn - m = d, n+m=49/dn + m = 49/d for each:

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

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

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.