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Difference between revisions of "1997 AIME Problems/Problem 1"

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== Solution ==
 
== Solution ==
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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.
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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> 0, 1 \pmod 4.</math> Thus, the answer is <math>500+250 = \boxed{750}.</math>
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The remaining <math>\boxed{750}</math> numbers, we can describe specific squares which fit the conditions:
 
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*For all odd <math>x = 2n+1</math>, <math>(n+1)^2 - (n^2) = x</math>.
 
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*For all <math>x = 4n</math>, <math>(n+1)^2 - (n-1)^2 = x</math>.
 
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 14:52, 2 March 2020

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 $0, 1 \pmod 4.$ Thus, the answer is $500+250 = \boxed{750}.$

See also

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
All AIME Problems and Solutions

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