Is there a number that when multiplied by $2$ equals a perfect square, and when multiplied by $3$ equals a perfect square?

Question

I am wondering if there is a number $n$ that when multiplied by $2$ and by $3$ equals a perfect square.

$$n\times 3 = k^2$$ $$n\times 2 = m^2$$

where $n$ is a whole number and not equal to $0$

I have tried every whole number up to $1000$, and my theory is that there isn't a number that fits the criteria. If I am wrong, please include the number that works for both, what it multiplies to, and the square roots of the products.

Answer 1

Multiply the first equation by 2 and the second by 3 to get

$6n = 2k^2=3m^2$.

There are an odd number of powers of 2 on the left side and an even number on the right, a contradiction.

Note that this works for $np=k^2, nq=m^2$ where $p$ and $q$ are distinct primes.

It can be generalized further, but that is left as an exercise.