Do there exist $m,n$ such that $6 = 2(2m+1)^2/(2n+1)^2$? Can the two numbers $n$ and $m$ exists such that
$$6=\frac{2(2m+1)^2}{(2n+1)^2}$$
where $\gcd(m,n) = 1$?
 A: By the comments above, it suffices to show that $\sqrt{3}$ is not a rational number. We proceed by contradiction. So suppose that $\sqrt{3}=\frac{a}{b}$ with $a,b$ integers such that $\text{gcd}(a,b)=1$. (We may assume this without loss of generality). Then $3=\frac{a^2}{b^2}$, hence $3a^2=b^2$. It follows that $3\mid b^2$ and thus that $3\mid b$. Thus we can write $b=3k$ for some integer $k$. Thus $3a^2=9k^2$, thus $a^2=3k^2$. In the same fashion we conclude that $3\mid a$, thus we may write $a=3l$ for some $l$. But then $\text{gcd}(a,b)\geq 3$ contrary to our assumption. Thus we conclude that $\sqrt{3}$ is irrational.
Let me give you an alternative proof. Notice that the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}$ is given by $x^2-3$. Indeed, clearly $\sqrt{3}$ is a root of this polynomial and by the criterium of Eisenstein, we see that this polynomial is irreducible. It follows that the extension degree  $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=2$. Thus $\left\{1,\sqrt{3}\right\}$ forms a $\mathbb{Q}$-basis of $\mathbb{Q}(\sqrt{3})$, it follows that $\sqrt{3}\notin \mathbb{Q}$. 
A: So you want to prove that without using the fact that $\sqrt{3}$ is irrational.
No problem.
Suppose that you have $3=\dfrac {(2m+1)^2}{(2n+1)^2}$.
Then $(2m+1)^2=3(2n+1)^2$.
That is, $4m^2+4m+1=12n^2+12n+3$.
So $4m^2+4m-12n^2-12n=2$ so the left hand side is divisible by $4$ but the right hand side is not, so such $m,n$ do not exist.
