1999 AIME Problems/Problem 3

Problem

Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square.Find the sum of all integers $n$ such that $\dfrac{12}{n}$ is also an integer.

Solution 1

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 $1+9+10+18=\boxed{38}$ .

Solution 2

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.

Solution 3

When $n \geq 12$ , we have $(n-10)^2 < n^2 -19n + 99 < (n-8)^2.$

So if $n \geq 12$ and $n^2 -19n + 99$ is a perfect square, then $n^2 -19n + 99 = (n-9)^2$

or $n = 18$ .

For $1 \leq n < 12$ , it is easy to check that $n^2 -19n + 99$ is a perfect square when $n = 1, 9$ and $10$ ( using the identity $n^2 -19n + 99 = (n-10)^2 + n - 1.)$

We conclude that the answer is $1 + 9 + 10 + 18 = \boxed{38}.$

1999 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1999 AIME Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

Solution 3

See also