Random Diophantine Equation

Let $\dfrac{n^3-2m^3}{6m} = x^2$ . This implies $n^3 - 2m^3 = 6m x^2 \implies n^3 = 2m^3 + 6mx^2= (m+x)^3 + (m-x)^3.$ By Fermats' Last Theorem, this has no integer solutions unless $m = x$ . However, $m = x$ implies $n^3 = 8m^3 \implies n = 2m$ , contradicting the fact that $m \neq \dfrac{n}{2}$ . Hence, the answer is $\boxed{0}$ .

Using Sreejato's notation below, if we let $X = 6n/m$ , $Y = 36x/m$ , we get $Y^2 = X^3-432$ . This is an elliptic curve with rank $0$ and torsion subgroup of order $3$ . So the only rational points are the point at infinity and $(12,\pm 36)$ , which leads to $x = \pm m, n = 2m$ as the only possible solutions.

(Obviously, the facts about the elliptic curve require some work to prove!)

Let $\frac{n^3-2m^3}{6m}=p^2$ . It implies that:

$n^3-2m^3=6mp^2 \Rightarrow n^3=6mp^2+2m^3$

The RHS must be even for all integers $(m,p)$ . Since the RHS even, the LHS must equal to the value of RHS and even. In other words, all positive integers for $n$ must be even. Thus, $n^3$ must be even too.

If $n$ is odd, then $n^3$ must be odd and the RHS must be odd and equal to the LHS. Since the odd number is not divisible by 2, then it contradicts both sides for the RHS is divisible by 2. Thus, there are no solutions for all odd positive integers $n$