For $x,y,n\in N$, find the smallest $n$ for which there exist four different pairs of solutions $(x,y)$ of $5x+6y=n$

Question

For $x,y,n\in N$, find the smallest $n$ for which there exist four different pairs of solutions $(x,y)$ of $5x+6y=n$

I found the following solution using a simple brute-force algorithm, but I have no idea how to find this answer using mathematical tools.

n = 101

1. $\quad x_1 =\,\, 1,\quad y_1 = 16$
2. $\quad x_2 = \,\,7,\quad y_2 = 11$
3. $\quad x_3 = 13,\quad y_3 = \,\,6$
4. $\quad x_4 = 19,\quad y_4 = \,\,1 $

Answer 1

If you look at how to get from one of your solutions to the next, note that $x$ increases by $6$ and $y$ decreases by $5$. In fact, for any $n$, with one solution $(x,y)$, all other solutions are given by $(x+6k, y-5k)$ for some integer $k$.

If all four solutions should have $x,y\geq1$, then because we want the minimal possible case, one solution ought to have $x=1$, and the solution at the other end ought to have $y=1$. Simple counting using the observation from the previous paragraph gives that $x=1$ must go together with $y=5\cdot3+1$, and $y=1$ must go together with $x=6\cdot3+1$.