Find all integers $x$ such that $x^2+3x+24$ is a perfect square. Find all integers $x$ such that $x^2+3x+24$ is a perfect square.  
My attempt:  
$x^2+3x+24=k^2$
$3(x+8)=(k+x)(k-x)$  
Now, do I find solution treating cases? But that doesn't seem very easy. Please help.

Complete the square to get $(x+3/2)^2 + 87/4$. We want this to be a square itself, so
$$(x+3/2)^2 + 87/4 = k^2.$$
This is the same as 
$$(2x+3)^2 + 87 = 4k^2.$$
Now consider the difference of squares $(2k - 2x - 3)(2k + 2x + 3) = 87 = 3\cdot 29$. Now there are only a few cases to check.


*

*$2k + 2x + 3 = 3$ and $2k - 2x - 3 = 29$ means $4k = 32$, $k = 8$.

*$2k + 2x + 3 = 87$ and $2k - 2x - 3 = 1$ means $4k = 88$, $k = 22$.


Do these give you any integer solutions for $x$?
