Hello,
I found a question in some old math's archive on the net which says,
Find the maximum possible integral value of $n$ such that $4n^2 + 5n - 8$ is the square of an integer.
Now if we put it in an equation, we get $$4n^2 + 5n - 8 = \alpha^2$$ where $\alpha^2$ is the square number.
Inserting this equation into the quadratic formula gives $$ \dfrac {-5 \pm \sqrt {5^2 - 4.4.-(8 + \alpha^2)}} {2.4}$$ and further simplification gives, $$ \dfrac {-5 \pm \sqrt {153 + 16\alpha^2}} 8$$ Now the value of the $\sqrt{153 + 16\alpha^2}$ should be odd because if it is then $-5 + \sqrt{153 + 16\alpha^2}$ will become an even number and might be divisible by $8$ giving an integral value for $n$.
One possible value for $\alpha^2$ is $1$, but I'm not sure whether it is the largest possible value.
So, i want to know if there is a way to find any other value for $\alpha$ such that $\sqrt{153 + 16\alpha^2}$ becomes an odd number.
ANY HELP IS APPRECIATED!!
Hint: $\;(2n+1)^2 = 4n^2+4n+1 \lt 4n^2 + 5n - 8 \lt 4n^2+8n+4 = (2n+2)^2\,$ for $\,n \gt 9\,$.