<s>[INST] 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 $$2n-1=25=5^2$$
 [/INST] $a^2-1=3b^2$ 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.
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