Find all integers $x$ such that $x^2+3x+24$ is a perfect square.
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.
Do these give you any integer solutions for $x$?
Best way is....
$$x^2 + 3x + 24 = k^2 \Rightarrow x^2 + 3x + 24-k^2 = 0\\$$
$$\triangle = b^2 - 4ac = 3^2 - 4(1)(24-k^2) = 9 - 96 + 4k^2 = 4k^2 - 87 = n^2\\$$
$$\to 4k^2 - n^2 = (2k-n)(2k+n) = 87 \\$$
$$\to 2k+n = 29, 2k-n = 3 \to 4k = 32 \to k = 8$$. Can you finish it?
Hint Consider $$4k^2=(2x+3)^2+87$$