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galactus · Mar 19, 2008, 01:40 PM

Here's another challenge problem in case anyone would like a try besides the mundane.

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

eeseely · Mar 21, 2008, 08:31 AM

Originally Posted by galactus

Here's another challenge problem in case anyone would like a try besides the mundane.

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

Using Microsoft Excel shows that the expression equaled a perfect square in the following sequence for n:

2 + 21 = 23 + 23 = 46 + 25 = 71 + 27 = 98 + 29 = 127... etc

This shows that the increment between integers has a definite pattern, i.e. each new increment is 2 more than the previous increment.

Based upon this observation, the sum of the integers resulting in the target formula equaling a perfect square would be infinite.

galactus · Mar 21, 2008, 02:42 PM

I am sorry, I am not following. But, I don't think there are infinite solutions. Here's how I done it.

Let $n^{2}-19n+99=k^{2}$

Then we have the quadratic equation $n^{2}-19n+(99-k^{2})=0$

By the quadratic formula, we get $n=\frac{19\pm{\sqrt{4k^{2}-35}}}{2}$

Since n is an integer, $4k^{2}-35$ must be a square, say it is $a^{2}$

Then we have $4k^{2}-35=a^{2}\Rightarrow{(2k)^{2}-a^{2}=35}$

We have two squares that differ by 35.

The only such squares are $(6^{2}, 1^{2}), \;\ (18^{2},17^{2})$

That is $k=3 , \;\ k=9$

If $k=3, \;\ n=\frac{19\pm{1}}{2}=10, \;\ 9$

If $k=9, \;\ \frac{19\pm{17}}{2}=18, \;\ 1$

The sum of the n's is $1+9+10+18=38$

eeseely · Mar 22, 2008, 03:03 PM

I made a drastic error when constructing my spreadsheet.

Here is the corrected spreadsheet results which validates your answer.

N N^2 -19N + 99 RESULT SQRT (RESULT)

1 1.000000 80.000000 81.000000 9.00
9 81.000000 -72.000000 9.000000 3.00
10 100.000000 -91.000000 9.000000 3.00
18 324.000000 -243.000000 81.000000 9.00