# Thread: For any integer x, 100 x^2 - 24 x - 379 is not a complete square

1. ## For any integer x, 100 x^2 - 24 x - 379 is not a complete square

Can $\displaystyle 100 x^2 - 24 x -379$ be a complete square for some integer x > 0?

Sorry I should add a condition x > 0 in the subject line.

2. Originally Posted by elim
Can $\displaystyle 100 x^2 - 24 x -379$ be a complete square for some integer x > 0?

Sorry I should add a condition x > 0 in the subject line.
Try $\displaystyle x=403$.

3. when x = 403, $\displaystyle 100x^2-24x-379=16230849$
$\displaystyle \sqrt{16230849}=4028.752784671\cdots$
Not work.

4. Originally Posted by elim
when x = 403, $\displaystyle 100x^2-24x-379=16230849$
$\displaystyle \sqrt{16230849}=4028.752784671\cdots$
Not work.
Whoops! I typed something in wrong earlier. My mistake!

5. Hello, elim!

Can $\displaystyle 100 x^2 - 24 x -379$ be a complete square for some integer x > 0?

If $\displaystyle 100x^2 - 24x - 379$ is the square of any real number,

. . then the equation: $\displaystyle x^2-24x - 379 \:=\:0$ has a discriminant equal to 0.

Does it?

6. Originally Posted by elim
Can $\displaystyle 100 x^2 - 24 x -379$ be a complete square for some integer x > 0?

Sorry I should add a condition x > 0 in the subject line.
$\displaystyle 100 x^2 - 24 x -379 = n^2\implies x=\frac{6\pm\sqrt{25n^2+9511}}{50}$

Thus we require $\displaystyle 25n^2+9511$ to be square to give $\displaystyle x$ a fighting chance to be an integer.

So it's equivalent to show the diophantine equation $\displaystyle (5n)^2-m^2=-9511$ has no solution.

7. Originally Posted by chiph588@
$\displaystyle 100 x^2 - 24 x -379 = n^2\implies x=\frac{6\pm\sqrt{25n^2+9511}}{50}$

Thus we require $\displaystyle 25n^2+9511$ to be square to give $\displaystyle x$ a fighting chance to be an integer.

So it's equivalent to show the diophantine equation $\displaystyle (5n)^2-m^2=-9511$ has no solution.
A solution to $\displaystyle (5n)^2-m^2=-9511$ is $\displaystyle (n,m)=(951,4756)$.

This implies $\displaystyle x=-95$... but you need $\displaystyle x>0$.

This is a really tough problem! What is this for?

8. We want to solve: $\displaystyle 100x^2-24x-379=n^2$

Complete the square by multiplying both sides by $\displaystyle 400$ and adding and subtracting $\displaystyle 24^2$: $\displaystyle (200x-24)^2-400n^2-152176=0$

Let $\displaystyle y=200x-24$: $\displaystyle y^2-400n^2-152176=0$

Factor: $\displaystyle (y+20n)(y-20n)=152176$

Now let's factor $\displaystyle 152176$: $\displaystyle 152176 = 2^4\cdot9511$

So you can see, we only have a finite number of cases to consider ($\displaystyle 20$ to be exact) when solving for $\displaystyle y$ and $\displaystyle n$
e.g. $\displaystyle \begin{cases} y+20n=4 \\ y-20n=38044 \end{cases}$ is one example.

I'll leave you to consider all these cases, but you should find that $\displaystyle y=19024$, or in other words $\displaystyle x=-95$, is the only time there is a solution in $\displaystyle \mathbb{Z}$.

Thus for $\displaystyle x>0$, $\displaystyle 100x^2-24x-379$ is never square.

9. Thanks chiph588@
I was asked to show that $\displaystyle 25x^2-6x-95=y(y-1)$ has no positive integer solutions.

And this leads to the problem here.

10. Originally Posted by elim
Thanks chiph588@
I was asked to show that $\displaystyle 25x^2-6x-95=y(y-1)$ has no positive integer solutions.

And this leads to the problem here.
How did you reduce this problem to $\displaystyle 100x^2-24x-379=n^2$?

11. solve y to get $\displaystyle y=\frac{1}{2} \left(1+\sqrt{100x^2-24x-379}\right)$

Then as you pointed, $\displaystyle x = (6+\sqrt{25n^2+9511})/50$

Thus $\displaystyle 25n^2+9511=m^2$ for some positive integer $\displaystyle m$. That is
$\displaystyle (m+5n)(m-5n)=9511$
Since 9511 is a prime, $\displaystyle m+5n = 9511, \quad m-5n = 1$
Thus $\displaystyle n=951$ but then $\displaystyle x$ is not an integer.

12. Originally Posted by elim
solve y to get $\displaystyle y=\frac{1}{2} \left(1+\sqrt{100x^2-24x-379}\right)$

Then as you pointed, $\displaystyle x = (6+\sqrt{25n^2+9511})/50$

Thus $\displaystyle 25n^2+9511=m^2$ for some positive integer $\displaystyle m$. That is
$\displaystyle (m+5n)(m-5n)=9511$
Since 9511 is a prime, $\displaystyle m+5n = 9511, \quad m-5n = 1$
Thus $\displaystyle n=951$ but then $\displaystyle x$ is not an integer.
This is better than my solution since there are much less cases to consider!