1997 AIME Problems/Problem 1

Problem

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

Solution

If we let the two squares be $a^2 - b^2 = x$ , then by difference of squares we have $(a-b)(a+b) = x$ . Notice that $a-b$ and $a+b$ have the same parities. This eliminates all numbers in the form of $4n+2$ : when $x=2(2n+1)$ is factored, one of the factors must be even, but not both, so its factors cannot have the same parity. However, one cannot be represented as the difference of squares.

For the remaining $\boxed{749}$ numbers with the exception of 1, we can describe specific squares which fit the conditions:

For all odd $x = 2n+1$ , $(n+1)^2 - (n^2) = x$ .
For all $x = 4n$ , $(n+1)^2 - (n-1)^2 = x$ .

1997 AIME (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1997 AIME Problems/Problem 1

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