Prove that an equation with two integer variables has no solution

Question

$$\dfrac{(2n)!}{((6k)n!(2^n)+1)} = n^2$$

How will you prove that it has no solution if $n$ and $k$ are integers?

Answer 1

It could be that I am mistaken, but:

For $n=0,1$, the claim is obviously true.

$(2n)!$ is divisible by $n^2$ (since it has factors $2n$ and $n$). So you can divide both sides by $n^2$, so that the right hand side becomes 1.

If $k\leq0$, then the assertion is clearly wrong. Otherwise we have an odd number that equals $\frac{(2n-1)!}{n}$. If $n\geq2$, this cannot hold (count the factors of 2 in $\frac{(2n-1)!}{n}$).

But again, it is quite late where I am now, and this may be not as easy. :)