if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ always square Let $n\in \mathbb{N}$, if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ is a perfect square.
For example $n=1$, it is clear.
and $n=13$ because  $$ 4\cdot 13^2-1=3\cdot 15^2$$ and $$2n-1=25=5^2$$
 A: $a^2-1=3b^2$ is a Pell equation, $a^2-3b^2=1$. The solutions correspond to
$$
a_k + b_k \sqrt3 = (2+\sqrt3)^{k}
$$
We have
$$
a_{k+2}=4a_{k+1}-a_k
, \quad
a_0 = 1
, \quad
a_1 = 2
$$
because $2+\sqrt3$ is a root of $x^2 - 4 x + 1$.
Therefore
$$
a_k \bmod 2 = 1,0,1,0,\dots
\\
a_k \bmod 3 = 1,2,1,2,\dots
$$
Write $4n^2-1=3m^2$ as $(2n)^2-3m^2=1$. Therefore, $2n=a_{2k+1}$ and so $2n \equiv 2 \bmod 3$.
Write $4n^2-1=3m^2$ as $(2n-1)(2n+1)=3m^2$. Since $3$ divides $2n+1$ and $2n-1$ and $2n+1$ are coprime, we must have that $2n-1$ is square.
A: Here I prove an explicit expression for $\sqrt{2n-1}:  \sqrt{\frac12(4n-3m+1)}+\sqrt{\frac32(4n-3m-1)}$
From lhf's answer, $a_{2k+1}+b_{2k+1}\sqrt3=(a_{2k}+b_{2k}\sqrt3)(2+\sqrt3)=(2a_{2k}+3b_{2k})+(a_{2k}+2b_{2k})\sqrt3$,
so $a_{2k+1}=2a_{2k}+3b_{2k}$ and $b_{2k+1}=a_{2k}+2b_{2k}$ and therefore $a_{2k}=2a_{2k+1}-3b_{2k+1}$. 
Also, from $a_{2k}^2-3b_{2k}^2=1$ we have $\sqrt{3(a_{2k}^2-1)}=3b_{2k}$.
Therefore, $\left(\sqrt{\frac12(a_{2k}+1)}+\sqrt{\frac32(a_{2k}-1)}\right)^2=\frac12(a_{2k}+1)+\frac32(a_{2k}-1)+\sqrt{3(a_{2k}^2-1)}$
$=2a_{2k}+b_{2k}-1=a_{2k+1}-1=2n-1.$
Note that $\sqrt{\frac12(a_{2k}+1)}$ and $\sqrt{\frac32(a_{2k}-1)}$ are integers,
by arguments similar to lhf's that $\sqrt{\dfrac{2n+1}3}$ and $\sqrt{2n-1}$ are.
A: An Experimental approach:
There are infinitely many integers like n such that:
$2n-1=k^2$; $k∈N$
They make following series:
$n=13, 41, 61, 181, 221 . . .$ 
therefore we may write:
$4\times 13^2-1=3\times 15^2$
$4\times 41^2-1=3\times 47.34^2$
$4\times 61^2-1=3\times 70.4^2$
$4\times 181^2-1=3\times 209^2$
$4\times 221^2-1=3\times 255.19^2$
.
.
For $4 . n^2-1=3 .m^2$ The condition $2n-1=k^2$ is necessary but not sufficient. That is if  equation ($4 . n^2-1=3 .m^2$) has integer solution then ($2n-1$) is definitely is a square. 
