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.

The remaining $\boxed{750}$ numbers, 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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