Well, by intuition, of course there is doesn't exist any nonzero integers, but how would you prove that? I was thinking of doing the GCD of $a$ and $b$ is $1$, but that leads me to nowhere.

  • $\begingroup$ Hint: Write $a=3^mn$ and $b=3^rs$ where $\gcd(3,n)=1=\gcd(3,s)$ $\endgroup$ – lulu May 8 '16 at 2:18
  • $\begingroup$ Hint: Use a proof by contradiction. If the square of a/b is indeed 3, what does that say about a and b? $\endgroup$ – Adrian Mungroo May 8 '16 at 2:23
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    $\begingroup$ The exponents of the prime factors of $a^2$ are all necessarily even while those of the prime $3$ in $3b^2$ are necessarily odd. Hence these numbers are distinct. $\endgroup$ – Piquito May 8 '16 at 2:23
  • $\begingroup$ @AdrianMungroo If the square of a/b is 3, then a and b are prime?? $\endgroup$ – nyorkr23 May 8 '16 at 2:26
  • $\begingroup$ @Piquito how do you know $a^2$ have even prime factors and $3b^2$ are odd? $\endgroup$ – nyorkr23 May 8 '16 at 2:27

It doesn't imply that they are prime. It means that a/b is then positive or negative root 3. Let's just take it to be positive. You now have a ratio of assumed integers giving you an irrational number, root 3. Do you think this makes sense? What conclusions can you make from this?

  • $\begingroup$ I intended this to be a comment, not an answer. My apologies. $\endgroup$ – Adrian Mungroo May 8 '16 at 2:35
  • $\begingroup$ Is OP allowed to assume that $\sqrt{3}$ is irrational here? $\endgroup$ – D_S May 8 '16 at 2:45
  • $\begingroup$ Yes indeed; it's like asking if you must assume 1 is an integer. $\endgroup$ – Adrian Mungroo May 11 '16 at 19:06
  • $\begingroup$ How so? If you are allowed to assume $\sqrt{3}$ is irrational, the problem is a complete triviality $\endgroup$ – D_S May 11 '16 at 21:57
  • $\begingroup$ Which it is. But if you want to go the extra mile for completeness, you may include a separate proof by contradiction showing the irrationality of sqrt3. $\endgroup$ – Adrian Mungroo May 11 '16 at 22:45

If $x$ is a rational number, and $x^2$ is an integer, then so is $x$. Do you see why?

Now if $a^2 = 3b^2$ for integers $a, b$, then $(\frac{a}{b})^2 = 3$, so $\frac{a}{b}$ is an integer. This implies that $3$ is the square of an integer, which it is not.

  • $\begingroup$ Couldn't root 3 squared equal to 3? I get what you're trying to say, but woudn't $(root 3)^2$ get us 3? $\endgroup$ – nyorkr23 May 8 '16 at 2:42
  • $\begingroup$ Well $\sqrt{3}$ is not a rational number :) , so no problem there. That is actually what you must argue here. $\endgroup$ – D_S May 8 '16 at 2:42
  • $\begingroup$ Wait, now I'm a bit confused. $a/b$ is suppose to be an integer, but from the question, we see that $a/b$ is not a integer, but a rational number. What am I not getting? $\endgroup$ – nyorkr23 May 8 '16 at 2:57
  • $\begingroup$ It is a priori a rational number, but it might also be an integer :) . For example, $221/13$ is at first glance merely a rational number, but it is in fact an integer. $\endgroup$ – D_S May 8 '16 at 4:46
  • $\begingroup$ Since $(\frac{a}{b})^2$ is equal to $3$, and $3$ is an integer, you have that $(\frac{a}{b})^2$ is an integer. And therefore... $\endgroup$ – D_S May 8 '16 at 4:46

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