2024 AMC 10A Problem 11
How many ordered pairs of integers satisfy ?
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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 ? It equals a square root, and a square root is never negative. So .
2. Get rid of the square root. Square both sides: , so
3. Factor. A difference of two squares factors: . So
4. List the ways to write as a product of two integers. Since , the second factor is at least as large as the first factor . The possibilities are:
| (average of the two) | |||
|---|---|---|---|
( is halfway between the two factors, and is half the gap between them.) The other orders, such as and , would give a negative , which is not allowed.
5. Check. , and . The negative values of give the same squares.
So there are pairs: .
The answer is .
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 may be negative and 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 is a square root, . Squaring, , so
The two factors are integers whose product is : the ordered factor pairs are . Solving , for each:
- : . Valid.
- : . Valid.
- : . Rejected ().
- : . Rejected.
- : . Valid.
- : . Valid.
Check: and . Four ordered pairs: .
The answer is .
Review in the AoPS books
- Introduction to Number Theory, Chapter 7: Algebra With Integers
- Introduction to Number Theory, Chapter 1: Integers: The Basics
- Introduction to Algebra, Chapter 11: Special Factorizations
- Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
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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.