https://artofproblemsolving.com/community/q1h557216p3238798

Since 12n^2 + 1 is a square number, we have 12n^2 = (a-1)(a+1) for some integer a. $a$ must be odd, let it be $2k+1.$ $(a-1)(a+1) = 2k \cdot (2k+2) = 12n^2 \Longrightarrow k(k+1) = 3n^2. $ However, note that $\gcd(k,k+1) = 1,$ hence, one of k,k+1 is 3b^2 and the other must be some square number c^2. Anyways, in terms of $k,$ we have $\sqrt{12n^2 + 1}= 2k+1.$ We must thus show that $k+1$ is the square of a number. Evidently, k+1 cant be 3b^2 since k would be a square which would leave it 2 mod 3. It follows that k+1 is a square of some integer c. edit: sniped

Let $\sqrt{12n^2+1}=x$. Take this to the form $$(x-1)(x+1) = 2^2\cdot 3\cdot n^2. $$Also, notice that $\gcd(x+1,x-1) =1/2$. We also have that $3\mid x+1$ or $3\mid x-1$. Case 1. $\gcd(x+1,x-1) =1$. We have $$\left(\frac{x\pm 1}{3}\right)\cdot(x\mp 1) = 2^2\cdot n^2. $$By gcd, we have therefore that $\left(\frac{x\pm 1}{3}\right)\ = r^2$ and $x\mp 1=q^2\implies x=q^2\pm 1$. Thus, $$q^2\pm 2=3r^2,$$also since $q$ or $r$ is even, we get a contradiction by modulo $4$. Case 2. $\gcd(x+1,x-1) =2$. We have $$\left(\frac{x\pm 1}{3}\right)\cdot(x\mp 1) = 2^2\cdot n^2. $$By gcd, we have therefore that $\left(\frac{x\pm 1}{3}\right)\ = 2r^2$ and $x\mp 1=2q^2\implies x=2q^2\pm 1$. Thus, $$q^2\pm 1=3r^2,$$note that $q^2+1=3r^2$ is impossible due to modulo $3$. We must have $q^2- 1=3r^2$, hence $x+1=2q^2$. We have $$ \sqrt{ \frac{\sqrt{12n^2+1}+1}{2}} =\sqrt{ \frac{x+1}{2}} = q\quad \square$$

$12n^2=(z-1)(z+1)=k$ for some $z$ then clearly as $4|k \implies 3n^2=\frac{z-1}{2}*\frac{z+1}{2}$ now as $gcd(\frac{z-1}{2},\frac{z+1}{2})=1$ so they both are simultaneously a perfect square and also if $3\mid \frac{z+1}{2}$ then $\frac{z-1}{2}\equiv 2\mod 3\neq$ perfect square. So $3|\frac{z-1}{2}$ hence $\sqrt{\frac{z+1}{2}}$ is Integer.$\blacksquare$