Regarding perfect squares Is there any positive integer $n > 2$ such that $(n - 1)(5n - 1)$ is a perfect square? It is observed that $(n - 1)(5n - 1)$ is of the form $4k$ or $4k+ 1$. 
Affirmative answers were given by Pspl and Mindlack (by providing some examples).
Now my question is the following:
Is there any characterization of positive integer $n$ such that $(n - 1)(5n - 1)$ is a perfect square?

If $m^2=(n-1)(5n-1)$, then $5m^2=5(n-1)(5n-1)=(5n-3)^2-4$.
Write this as $x^2-5y^2=4$, for $x=5n-3$ and $y=m$.
Write this as $N\left(\frac{x+y\sqrt5}{2}\right)=1$. Then $\frac{x+y\sqrt5}{2}=\left(\frac{1+\sqrt5}{2}\right)^{2k}$, since $\frac{1+\sqrt5}{2}$ is a fundamental unit of norm $-1$.
Thus, $\frac{x+y\sqrt5}{2}=\left(\frac{3+\sqrt5}{2}\right)^{k}$. We just need to find those $x$ such that $x \equiv -3 \equiv 2 \bmod 5$.
Write $\frac{x_k+y_k\sqrt5}{2}=\left(\frac{3+\sqrt5}{2}\right)^{k}$. Then $x_{k+1}=\frac{3x_k+5y_k}{2}$ and $y_{k+1}=\frac{x_k+3y_k}{2}$, with $x_0=2, y_0=0$.
Then, $x_{k+2}=\frac{7x_k+15y_k}{2}$ and $y_{k+2}=\frac{3x_k+7y_k}{2}$. This gives $x_{k+2} \equiv x_k \bmod 5$. Thus $x_{2k} \equiv 2 \bmod 5$, since $x_0 = 2$.
Thus, the solutions of $m^2=(n-1)(5n-1)$ are exactly $n=\frac{x_{2k}+3}{5}$ and $m=y_{2k}$.
Equivalently, $n_k=\frac{u_{k}+3}{5}$ and $m=v_{k}$, where $\frac{u_k+v_k\sqrt5}{2}=\left(\frac{1+\sqrt5}{2}\right)^{4k}=\left(\frac{7+3\sqrt5}{2}\right)^{k}$, which gives $u_{k+1}=\frac{7u_k+15v_k}{2}$ and $v_{k+1}=\frac{3u_k+7v_k}{2}$, with $u_0=2, v_0=0$.
