Welcome to our community

Be a part of something great, join today!

Find the number of positive integer values

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,753
Hi MHB,

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!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hint: Try using:

\(\displaystyle n=-2k^2-689k\) where \(\displaystyle k\in\mathbb{Z}\).

Can you show then that one of the roots must be $x=k$?

And then solve \(\displaystyle n>0\) to see how many integers you get.
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,753
Hint: Try using:

\(\displaystyle n=-2k^2-689k\) where \(\displaystyle k\in\mathbb{Z}\).

Can you show then that one of the roots must be $x=k$?

And then solve \(\displaystyle n>0\) to see how many integers you get.
Hi MarkFL,

Thanks for helping me out again!

Now everything makes perfect sense to me...the problem told us $n$ is positive integer, and we could rewrite the given quadratic equation to make $n$ the subject and then set its equivalent greater than zero...

So from

\(\displaystyle n=-2k^2-689k\) where \(\displaystyle k\in\mathbb{Z}\),

we have

\(\displaystyle n=k(-2k-689)\)

and since the quadratic equation has only one integer root, then $x=k$ must be right and now, if we set $n>0$, we need to consider for two cases. First case deals with the case where $k<0$ and second be the case where $k>0$.

If $k<0$, in order for $n>0$, we see that we must set $-2k-689<0$ or $k>-344.5$.

The number of positive $n$ values that we can get here is hence 344.

The second case gives us zero solution. Therefore, the total number of positive integer values of $n$ such that the equation $2x^2+689x+n=0$ have an integer solution is 344.

Thank you Mark for your help!
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, I get the same number of values. I found with the value for $n$ I suggested, we get:

\(\displaystyle f(x)=(x-k)\left(2(x+k)+689 \right)=0\)

The first factor gives us the integral root:

\(\displaystyle x=k\)

While the second root is:

\(\displaystyle x=-\left(k+\frac{689}{2} \right)\)

which cannot be an integer for any value of $k$.

In order for $n$ to be a positive integer, we require:

\(\displaystyle 2k^2+689k<0\)

\(\displaystyle k(2k+689)<0\)

Thus:

\(\displaystyle -344\le k\le-1\)

And, as you found, there are 344 values $n$ may have to satisfies the given requirements.
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
solution by MARKFL and anemone both good

here is another

let (2x+ a)(x+b) = 2x^2+689x+n

as n is +ve both a and b positive or -ve,

a is odd else 2 integer solutions


or 2x^2 + (a + 2b) = 2x^2+689x+n

a and b cannot - ve as sum is positive

0 < 2b < 689 and hence 0 < b < 344.5

so there are 344 values of b and also n