Prove that there are no non-zero integers $m$ and $n$ such that $m^2 = 180 n^4$

Question

I made the following:

$(m^2)^{1/2} =(n^4)^{1/2} 180^{1/2} $

Then $|m| = (n^2) 180^{1/2} $ but $180^{1/2} = 6 (5^{1/2})$ isn't an integer number. Then $m$ is an integer number if and only if $n= a (5^{1/2})$, $a$ is any integer number. That is, if $m$ is an integer number then $n$ isn't an integer and vice versa

Then, there aren't non-zero integers $m$ and $n$ such that $m^2 = 180 n^4$

Is right my demonstration? Is there any more clearly form to prove this exercise?

Answer 1

For $m^2 = 180 n^4$, assuming there is an integer solution, implies that $m^2$ is wholly divisible by $180 = 2\cdot 2\cdot 3\cdot 3\cdot 5$.

That is, $\frac{m^2}{2\cdot 2\cdot 3\cdot 3\cdot 5} = n^4$

Now, factors of $m$ must appear as an even quantity of each in $m^2$. So if we divide $m^2$ by a single factor of $5$ we will be left with an odd number of 5 factors which contradicts $n$ being a $4th$ power.