TITLE: $n^3+7$ cannot be a square number.
QUESTION [7 upvotes]: Let $n$ be a positive integer. What I have to show is that $n^3+7 \neq k^2$ for any integer $k$.
I assumed that $n^3+7=k^2$ for some integer $k$.
What I did :
$$(n+2)(n^2-2n+4)=k^2+1$$
$$k^2=-1 \mod{n+2}$$
$$(\frac{k}{n+2})=1$$
Another approach.
$$n^3+7=k^2 \mod{4} $$
Then $n$ should be a form of $4u+1$ and $k$ should be a form of $4l$ or $4l+2$.
But I failed to proceed this approach because it is too complicate.

REPLY [2 votes]: This is rather a classic problem regarding a specific type of Diophantine Equation called 'Mordell's equation', which is $y^2=x^3+k (x\in\mathbb{Z})$
Take a look at this  article by Keith Conrad. Theorem 2.1. will solve your problem.