1999 AIME Problems/Problem 3

Revision as of 16:27, 13 May 2012 by Hydroxide (talk | contribs) (Solution)

Problem

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

Solution

If the perfect square is represented by $x^2$, then the equation is $n^2 - 19n + 99 - x^2 = 0$. The quadratic formula yields

$n = \frac{19 \pm \sqrt{361 - 4(99 - x^2)}}{2}$

In order for this to be an integer, the discriminant must also be a perfect square, so $4x^2 - 35 = q^2$ for some nonnegative integer $q$. This factors to

$(2x + q)(2x - q) = 35$

$35$ has two pairs of positive factors: $\{1,\ 35\}$ and $\{5,\ 7\}$. Respectively, these yield $9$ and $3$ for $x$, which results in $n = 1,\ 9,\ 10,\ 18$. The sum is therefore $\boxed{038}$.

We have

$n^2-19n+99=x^2\Longleftrightarrow n^2-19n+99-x^2=0\Longleftrightarrow n=\frac{19\pm \sqrt{4x^2-35}}{2}$

This equation has solutions in integers if and only if for some odd nonnegative integer $q$, $4x^2-35=q^2$, or $(2x+q)(2x+q)=35$. Because $q$ is odd, this makes both of the factors $2x+q$ and $2x-q$ odd,

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.

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