Find all positive integers solutions for $x^2-xy-y^2=1$

Question

Find all positive integers solutions 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)$$ y^2+xy-x^2+1=0$$ $$D=x^2+4x^2-4=5x^2-4=m^2, m\in \mathbb Z$$

Answer 1

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: $(13,8), (233,144), (4181, 2584), ...$

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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 $(y_1, x_1, y_2, x_2, y_3, x_3, ...) = (1, 2, 3, 5, 8, 13, ...)$..

Answer 2

I wrote a recent program to do this:

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental_A
2  1
1  1


  3^2 - 5 1^2 = 4

1 x^2 + -1 x y -1 y^2 = 1

Fri Apr 29 10:40:50 PDT 2016

x:  2  y:  1 ratio: 2  seed 
x:  5  y:  3 ratio: 1.666666666666667
x:  13  y:  8 ratio: 1.625
x:  34  y:  21 ratio: 1.619047619047619
x:  89  y:  55 ratio: 1.618181818181818
x:  233  y:  144 ratio: 1.618055555555555
x:  610  y:  377 ratio: 1.618037135278515
x:  1597  y:  987 ratio: 1.618034447821682
x:  4181  y:  2584 ratio: 1.618034055727554
x:  10946  y:  6765 ratio: 1.618033998521803
x:  28657  y:  17711 ratio: 1.618033990175597
x:  75025  y:  46368 ratio: 1.618033988957902
x:  196418  y:  121393 ratio: 1.618033988780243
x:  514229  y:  317811 ratio: 1.618033988754322
x:  1346269  y:  832040 ratio: 1.618033988750541
x:  3524578  y:  2178309 ratio: 1.618033988749989
x:  9227465  y:  5702887 ratio: 1.618033988749908

Fri Apr 29 10:41:10 PDT 2016
2  1
1  1
Inverse of given automorphism of quadratic form:
1  -1
-1  2
jagy@phobeusjunior:~$ 

As you can see, the $x$ values are the odd index Fibonacci numbers, and the $y$ values are the even index Fibonaccis. The formal way to get from one solution $(x,y)$ to the next larger is $$ (x,y) \mapsto (2x+y, x+y). $$ As the determinant of the automorphism matrix is $1$ and the trace $3,$ we use Cayley-Hamilton to get separate $$ x_{n+2} = 3 x_{n+1} - x_n, $$ $$ y_{n+2} = 3 y_{n+1} - y_n. $$

The same information is contained and explained in Conway's Topograph diagram for this quadratic for $x^2 - xy- y^2.$ I drew this picture for my answer at Solve the following equation for x and y: where it was used for $x^2 - xy - y^2 = 5.$ In today's problem we found $x^2 - xy - y^2 = 1.$