I want to find a pair of integers $X$, $Y$ which satisfy $X^2 - 2Y^2=1$ such that $X > Y > 50$

Question

Find a pair of integers $X, Y$, which satisfy $X^2 - 2Y^2 = 1$, such that $X > Y > 50$.

I have started by finding a pair of much smaller integers that work: $X(1) = 3$ and $Y(1) = 2$.

When I looked up a solution it was as follows: $$\begin{array} &X(2) = 3 × 3 + 4 × 2 = 17, & Y(2) = 2 × 3 + 3 × 2 = 12 \\ X(3) = 3 × 17 + 4 × 12 = 99, & Y(3) = 2 × 17 + 3 × 12 = 70 \end{array}$$ So $X = 99$ and $Y = 70$ is such a pair.

But what is the method used? I tried searching for similar questions, but didn't find an answer to this.

Answer 1

$$X_{i+1}^2-2Y_{i+1}^2 = (3X_i + 4Y_i)^2 - 2(2X_i + 3Y_i)^2 =$$ $$ = (9-2\cdot4)X_i ^ 2 + (12-12)X_i \cdot Y_i + (16-2\cdot9)Y_i ^ 2= $$ $$= X_{i}^2-2Y_{i}^2 = 1$$

Answer 2

Another (more naive) way of looking at this: we want all possible $k$ for which $(Y+k)^2-2Y^2=1$. Solving this for $Y$ (in terms of $k$) we get $$ Y=k\pm\sqrt{2k^2-1} $$ We will want to ignore the minus, since our solutions must be positive and we only want the solutions for when $\sqrt{2k^2-1}$ is an integer.

By inspection we get the first 4 solutions: $$ \begin{array}{c|c|c|c} k & \sqrt{2k^2-1}& Y=k+\sqrt{2k^2-1} & X=Y+k \\ \hline 1 & 1 & 2 & 3 \\ 5 & 7 & 12 & 17 \\ 29 & 41 & 70 & 99 \\ 169 & 239 & 408 & 577 \\ \end{array} $$