Difference between revisions of "1997 AIME Problems/Problem 1"
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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. | + | 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. |
| − | + | 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 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>. | *For all <math>x = 4n</math>, <math>(n+1)^2 - (n-1)^2 = x</math>. | ||
Revision as of 21:02, 4 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
If we let the two squares be 
, then by difference of squares we have 
. Notice that 
and 
have the same parities. This eliminates all numbers in the form of 
: when 
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 
numbers with the exception of 1, we can describe specific squares which fit the conditions:
- For all odd
, 
. - For all
, 
.
, 
.
, 
.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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
