How many positive integers $n$ are there such that $2n$ and $3n$ both perfect squares?

Question

How many positive integers $n$ are there, such that both $2n$ and $3n$ are perfect squares? I tried to use modular arithmetic, but I'm stuck.

Answer 1

If $x\ne0,$

$$2x=a^2,3x=b^2\implies \frac{a^2}{b^2}=\frac23\iff \left(\frac ab\right)^2=\frac23$$

or on multiplication, $$6x^2=a^2b^2\iff \left(\frac{ab}x\right)^2=6$$

which is impossible as $a,b,x$ are integers

Answer 2

If you factor a perfect square, you'll see that every prime has an even power. If $2n$ has an even power of $2$, then $n$ has an odd power. Now consider the power of $2$ in $3n$. Since $2$ and $3$ are (co)prime, the power of $2$ in $3n$ is the power of $2$ in $n$, thus odd. Therefore, $3n$ can't be a perfect square.

Answer 3

Let $x=r\cdot2^a\cdot3^b$ where $(r,6)=1$

As $2x$ is perfect square, $r\cdot2^{a+1}\cdot3^b$ must be, so $r$ must be perfect square $=s^2$(say) and $a+1,b$ each must be even $\implies a$ must be odd

As $3x$ is perfect square, $s^2\cdot2^a\cdot3^{b+1}$ must be, so $a,b+1$ each must be even $\implies b$ must be odd

So, we reach a contradiction for the parity of $a,b$