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?

As $(n-1)(5n-1)=m^2$ for some $m\in\mathbb{N}$, then
$$\begin{split}(3n-1)^2-(2n)^2&=m^2\\m^2+(2n)^2&=(3n-1)^2\end{split}$$
By the formula for Pythagorean Triple, we can use it.
Let $m=x^2-y^2, 2n=2xy, 3n-1=x^2+y^2$ and assume $x>y>0$. Then
$$\begin{cases}
(x + y)^2 =5n-1 \\
(x-y)^2 = n-1
\end{cases}$$
By the assumption, $x+y,x-y>0$. So $x=\frac{\sqrt{5n-1}+\sqrt{n-1}}{2},y=\frac{\sqrt{5n-1}-\sqrt{n-1}}{2}$.
$\because x,y\in\mathbb{N}$
$\therefore n-1,5n-1$ must be perfect squares in order to have $(n-1)(5n-1)$ be a perfect square.
