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- Feb 14, 2012

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This problem drives me crazy because first, I have not encountered a problem like this before (but this is only an excuse and it must be my incompetence that holds me back from cracking it successfully) and I called off the attempt because I don't think I can solve it.

Problem:

For how many positive integer values of n does the equation $2x^2+689x+n=0$ have an integer solution?

Attempt:

roots$=\dfrac{-689 \pm \sqrt{689^2-8n}}{4}=\dfrac{-689 \pm k}{4}$ where $k=\sqrt{689^2-8n}$ hence

roots$=-172.25+0.25k, -172.25-0.25k$

roots$=-172.25+m-0.25, -172.25-(m-0.25)$ where $m-0.25=0.25k$

roots$=-172.5+m, -172-m$

In order for the original given quadratic equation to have only one integer solution, we see that $m$ must be an integer.

If we work on the product of the roots, we see that

$(-172.5+m)(-172-m)=\dfrac{n}{2}$

which simplifies to

$59340+m-2m^2=n$

Okay, if I let $m=150$ I then get $n=14490$ but I don't believe I have to try out all the possible values of $m$ for this problem.

And I don't know what else I can do to determine the number of positive integers $n$ that the problem asks right from where I stopped.

Any help/advice would be greatly appreciated. Thanks!