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.
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.
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?
$\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.
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.
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.