Regarding perfect squares

Question

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?

Answer 1

The following are solutions: $n= 10, 65, 442, 3026, 20737, 142130, 974170.$.

Answer 2

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.

Answer 3

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 $u_0=2, v_0=0$.