Difference between revisions of "1999 AIME Problems/Problem 3"
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:<math> n = \frac{19 \pm \sqrt{361 - 4(99 - x^2)}}{2}</math> | :<math> n = \frac{19 \pm \sqrt{361 - 4(99 - x^2)}}{2}</math> | ||
| − | In order for this to be an [[integer]], the [[discriminant]] must also be a perfect square, so <math> | + | In order for this to be an [[integer]], the [[discriminant]] must also be a perfect square, so <math>4x^2 - 35 = q^2</math> for some [[nonnegative]] integer <math>q</math>. This [[factoring | factors]] to |
| − | :<math>( | + | :<math>(2x + q)(2x - q) = 35</math> |
| − | <math>35</math> has two pairs of positive [[divisor | factors]]: <math>\{1,\ 35\}</math> and <math>\{5,\ 7\}</math>. Respectively, these yield | + | <math>35</math> has two pairs of positive [[divisor | factors]]: <math>\{1,\ 35\}</math> and <math>\{5,\ 7\}</math>. Respectively, these yield <math>9</math> and <math>3</math> for <math>x</math>, which results in <math>n = 1,\ 9,\ 10,\ 18</math>. The sum is therefore <math>038</math>. |
== See also == | == See also == | ||
Revision as of 16:47, 3 July 2007
Problem
Find the sum of all positive integers 
for which 
is a perfect square.
Solution
If the perfect square is represented by 
, then the equation is 
. The quadratic formula yields
In order for this to be an integer, the discriminant must also be a perfect square, so 
for some nonnegative integer 
. This factors to
has two pairs of positive factors: 
and 
. Respectively, these yield 
and 
for 
, which results in 
. The sum is therefore 
.
See also
| 1999 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
