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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.

Michael Hardy
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nyorkr23
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  • Hint: Write $a=3^mn$ and $b=3^rs$ where $\gcd(3,n)=1=\gcd(3,s)$ – lulu May 08 '16 at 02:18
  • Hint: Use a proof by contradiction. If the square of a/b is indeed 3, what does that say about a and b? – Adrian Mungroo May 08 '16 at 02:23
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    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. – Piquito May 08 '16 at 02:23
  • @AdrianMungroo If the square of a/b is 3, then a and b are prime?? – nyorkr23 May 08 '16 at 02:26
  • @Piquito how do you know $a^2$ have even prime factors and $3b^2$ are odd? – nyorkr23 May 08 '16 at 02:27
  • Not even prime factors but even exponents of the prime factors: the square of $p^n$ is equal to $p^{2n}$ for each prime factor $p$ and $a^2=(\prod p_1^{k_1}.....p_n^{k_n})^2=\prod p_1^{2k_1}.....p_n^{2k_n}$. Hence for $3b^2$ the exponent of $3$ will be odd. – Piquito May 08 '16 at 22:58

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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?

Adrian Mungroo
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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.

D_S
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  • 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? – nyorkr23 May 08 '16 at 02:42
  • Well $\sqrt{3}$ is not a rational number :) , so no problem there. That is actually what you must argue here. – D_S May 08 '16 at 02:42
  • 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? – nyorkr23 May 08 '16 at 02:57
  • 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. – D_S May 08 '16 at 04:46
  • 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... – D_S May 08 '16 at 04:46
  • $(a/b)^2$ is an integer, then $a/b$ is √3, which isn't a rational number nor an integer...oh which means it's impossible for $a$ and $b$ to be integers because there aren't any integers of $a/b$ ratio that could create √3 ? Am i correct? – nyorkr23 May 08 '16 at 04:51
  • How do you know that $\sqrt{3}$ cannot be expressed as a ratio of integers? I think that is basically what you have to prove. I have given you a hint on how to do that (basically, if $\sqrt{3}$ is rational, then it has to be an integer). – D_S May 08 '16 at 04:52
  • Aha, I have figured how to show that $√3$ is an irrational since if prime $p$, which is 3 in this case, the rational of p is irrational. Then using what you have suggested, I reworked the problem where I found that 3 could divide into $a^2$ and that 3 also could also divide into $b$, but since $a/b$ should be in the simplest form and both $a$ and $b$ are suppose to be coprimes, the GCD of them should be $1$, not $p$, therefore the contradiction. Now that I've proved it and gotten the answer...how would I start answering the question in order to get to these steps? – nyorkr23 May 08 '16 at 05:05
  • I think you are making this too complicated. Why not follow the steps I gave in my answer, it leads you directly to the solution. – D_S May 08 '16 at 15:40