Difference between revisions of "1999 AIME Problems/Problem 3"
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| − | :<math>n^2-19n+99=x^2\ | + | :<math>n^2-19n+99=x^2\iff n^2-19n+99-x^2=0\iff n=\frac{19\pm \sqrt{4x^2-35}}{2}</math> |
This equation has solutions in integers if and only if for some odd nonnegative integer <math>q</math>, <math>4x^2-35=q^2</math>, or <math>(2x+q)(2x-q)=35</math>. Because <math>q</math> is odd, this makes both of the factors <math>2x+q</math> and <math>2x-q</math> odd, so <math>(2x+q, 2x-q)=(35, 1)</math> or <math>(7,5)</math>, giving <math>x=3</math> or <math>9</math>. This gives <math>n=1, 9, 10,</math> or <math>18</math>, and the sum is <math>1+9+10+18=\boxed{038}</math>. | This equation has solutions in integers if and only if for some odd nonnegative integer <math>q</math>, <math>4x^2-35=q^2</math>, or <math>(2x+q)(2x-q)=35</math>. Because <math>q</math> is odd, this makes both of the factors <math>2x+q</math> and <math>2x-q</math> odd, so <math>(2x+q, 2x-q)=(35, 1)</math> or <math>(7,5)</math>, giving <math>x=3</math> or <math>9</math>. This gives <math>n=1, 9, 10,</math> or <math>18</math>, and the sum is <math>1+9+10+18=\boxed{038}</math>. | ||
Revision as of 15:32, 13 May 2012
Problem
Find the sum of all positive integers 
for which 
is a perfect square.
Solution
We have
This equation has solutions in integers if and only if for some odd nonnegative integer 
, 
, or 
. Because is odd, this makes both of the factors 
and 
odd, so 
or 
, giving 
or 
. This gives 
or 
, and the sum is 
.
Alternate Solution
Suppose there is some 
such that 
. Completing the square, we have that 
, that is, 
. Multiplying both sides by 4 and rearranging, we see that 
. Thus, 
. We then proceed as we did in the previous solution.
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 | ||
