The smallest $n$ for which $19n+1$ and $95n+1$ are perfect squares

Question

Find the smallest possible integer $n$ for which $19n+1$ and $95n+1$ are both perfect squares.

I somehow managed to show that $n$ is odd but couldn't find any solution for which both of them are perfect squares.

Answer 1

Assume,

$$95n+1 = x^2$$

$$19n+1= y^2$$

Eliminate $n$ between them and you get the Pell equation,

$$x^2-5y^2 = -4$$

I assume you know how to find all solutions $x,y$. It is then a simple matter to test them such that,

$$n = \frac{x^2-1}{95} = \frac{y^2-1}{19}$$

is an integer. The smallest ones are $n= 134232,\,920040,\,4481374227696,\, 30715795905552,\dots$ ad infinitum. (There doesn't seem to be odd n.)

Equivalently, for $m \ge 0$,

$$n = \frac{F_{18m+17}^2-1}{19} = 134232,\,4481374227696,\dots$$

$$n = \frac{F_{18m+19}^2-1}{19} = 920040,\,30715795905552,\dots$$

where $F_m$ is a Fibonacci number.