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.

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    $\begingroup$ What does "integers (+ve) - x" mean? $\endgroup$ – Pockets Jun 8 '14 at 5:47
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    $\begingroup$ @SamuelLijin, it means "positive". $\endgroup$ – Karolis Juodelė Jun 8 '14 at 5:52
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    $\begingroup$ Really? Is it really shorter/more comfortable/faster to write "(+ve)-" than "positive"? Because it obviously isn't clearer. $\endgroup$ – DonAntonio Jun 8 '14 at 11:16

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.


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

  • $\begingroup$ Thanks. I too came to a similar situation approximately. But the answer says "1". Is that wrong? $\endgroup$ – Krishnaar Jun 8 '14 at 6:25
  • $\begingroup$ @Krishnaar, Have you noticed, my assumption in the first line? (Though you say $x>0$) $\endgroup$ – lab bhattacharjee Jun 8 '14 at 6:29
  • $\begingroup$ @ lab bhattacharjee- sorry, I didn't. But thanks for pointing it out again. This is yet another silly folly of mine. $\endgroup$ – Krishnaar Jun 8 '14 at 6:31
  • $\begingroup$ @Krishnaar, Also check my other answer $\endgroup$ – lab bhattacharjee Jun 8 '14 at 6:32
  • $\begingroup$ @ lab bhattacharjee-thats even more cool! like an olympiad proof :) $\endgroup$ – Krishnaar Jun 8 '14 at 6:36

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 $a,b$

  • $\begingroup$ @Downvoter, Please pinpoint the mistake. Also, have a look into the comment of OP in my other answer regarding this. $\endgroup$ – lab bhattacharjee Jun 8 '14 at 8:45

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