Find all positive inegers solution for $x^2-xy-y^2=1$ 
Find all positive inegers solution for the following diophantine equation 
  $$x^2-xy-y^2=1$$

My work so far
1)$$x^2-xy-y^2-1=0$$
$$D=y^2+4(y^2+1)=5y^2+4=k^2, k \in \mathbb Z$$
2)$$ y^2+xy-x^2+1=0$$
$$D=x^2+4x^2-4=5x^2-4=m^2, m\in \mathbb Z$$
 A: You reached $k^2-5y^2=4$.  To proceed from here, you need the theory of Pell equations.  One way is to consider the LHS which has a factorization in $\mathbb Z[\sqrt5]$.  
In $\mathbb Z[\sqrt5]$, the norm is $N[a+b\sqrt5] = a^2-5b^2$, so we are actually seeking elements s.t. $N[z]=4$.  Clearly $z_0=3+\sqrt5$ is a solution, but there are infinite units satisfying $N[u]=1$, so there are a lot more solutions of form $z_0 u^n$.
Case 1: $z_0 = 3+\sqrt5, u=9\pm\sqrt5$
Here $z_n = (3+\sqrt5)(2+\sqrt5)^{2n}=(3+\sqrt5)(9\pm 4\sqrt5)^n$ is always a solution for $(k, \pm y)$.
Case 2: $z_0 = 2, u=9+4 \sqrt5$
Here again $z_n =2(9+ 4\sqrt5)^n$ generates solutions for $(k, y)$.
The first few solutions using 
Case 1 positive branch is $(x, y) = (2, 1), (34, 21), (610, 377), (10946, 6765), (196418, 121393)...$.  
Case 1 negative branch is $(5, 3), (89, 55), (1597, 987), ...$
Similarly from Case 2 we have: $(13,8), (233,144), (4181, 2584), ...$ 

P.S. If you prefer the recursion, it is $f(n+2)=18f(n+1)-f(n)$, for both $x_n, y_n$, ...
or a simpler one can be observed from the pattern - all Fibonacci numbers are covered in the form $(y_1, x_1, y_2, x_2, y_3, x_3, ...) = (1, 2, 3, 5, 8, 13, ...)$..
