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

See also

1999 AIME (ProblemsAnswer KeyResources)
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

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