Art of Problem Solving

1997 AIME Problems/Problem 1: Difference between revisions

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== Solution ==
== Solution ==
If we let the two squares be <math>a^2 - b^2 = x</math>, then by difference of squares we have <math>(a-b)(a+b) = x</math>. Notice that <math>a-b</math> and <math>a+b</math> have the same [[parity|parities]]. This eliminates all numbers in the form of <math>4n+2</math>: when <math>x=2(2n+1)</math> 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.
Notice that all odd numbers can be obtained by using <math>(a+1)^2-a^2=2a+1,</math> where <math>a</math> is a nonnegative integer. All multiples of <math>4</math> can be obtained by using <math>(b+1)^2-(b-1)^2 = 4b</math>, where <math>b</math> is a positive integer. Numbers congruent to <math>2 \pmod 4</math> cannot be obtained because squares are <math>-1, 0, 1 \pmod 4.</math> Thus, the answer is <math>500+250 = \boxed{750}.</math>
 
For the remaining <math>\boxed{749}</math> numbers with the exception of 1, we can describe specific squares which fit the conditions:
*For all odd <math>x = 2n+1</math>, <math>(n+1)^2 - (n^2) = x</math>.
*For all <math>x = 4n</math>, <math>(n+1)^2 - (n-1)^2 = x</math>.


== See also ==
== See also ==

Revision as of 03:01, 10 January 2019

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

Notice that all odd numbers can be obtained by using $(a+1)^2-a^2=2a+1,$ where $a$ is a nonnegative integer. All multiples of $4$ can be obtained by using $(b+1)^2-(b-1)^2 = 4b$, where $b$ is a positive integer. Numbers congruent to $2 \pmod 4$ cannot be obtained because squares are $-1, 0, 1 \pmod 4.$ Thus, the answer is $500+250 = \boxed{750}.$

See also

1997 AIME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.