Difference between revisions of "1999 AIME Problems/Problem 3"

Revision as of 18:39, 4 July 2013

Problem

Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square.

Solution

If $n^2-19n+99=x^2$ for some positive integer $x$ , then rearranging we get $n^2-19n+99-x^2=0$ . Now from the quadratic formula,

$n=\frac{19\pm \sqrt{4x^2-35}}{2}$

Because $n$ is an integer, this means $4x^2-35=q^2$ for some nonnegative integer $q$ . Rearranging gives $(2x+q)(2x-q)=35$ . Thus $(2x+q, 2x-q)=(35, 1)$ or $(7,5)$ , giving $x=3$ or $9$ . This gives $n=1, 9, 10,$ or $18$ , and the sum is $1+9+10+18=\boxed{038}$ .

Alternate Solution

Suppose there is some $k$ such that $x^2 - 19x + 99 = k^2$ . Completing the square, we have that $(x - 19/2)^2 + 99 - (19/2)^2 = k^2$ , that is, $(x - 19/2)^2 + 35/4 = k^2$ . Multiplying both sides by 4 and rearranging, we see that $(2k)^2 - (2x - 19)^2 = 35$ . Thus, $(2k - 2x + 19)(2k + 2x - 19) = 35$ . We then proceed as we did in the previous solution.

@@ Line 1: / Line 1: @@
 == Problem ==
-Find the sum of all [[positive integer]]s <math>\displaystyle n</math> for which <math>\displaystyle n^2-19n+99</math> is a [[perfect square]].
+Find the sum of all [[positive integer]]s <math>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]].
 == Solution ==
@@ Line 17: / Line 17: @@
 [[Category:Intermediate Number Theory Problems]]
 [[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

1999 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1999 AIME Problems/Problem 3"

Revision as of 18:39, 4 July 2013

Contents

Problem

Solution

Alternate Solution

See also