positive integer solution for $x^2-51y^2=1$? 
Find all positive integer solutions of: $$x^2-51y^2=1$$

I'm a bit lost. Tried to massage it into $(x-10y)(x+10y)=(1-7y)(1+7y)$ but didn't find anything fruitful.
 A: This Pellian equation has infinitely many solutions. The general method of solution is as follows.
Find the first convergent of the continued fraction fraction expansion of $\sqrt 51$. Since $7^2 < 51 < 8^2$, the first convergent or approximation of $\sqrt 51$ is $50/7$.
Find the smallest solution $(x_1,y_1)$ which is given by 
$x_1$ = numerator of the first convergent of the continued fraction fraction expansion of $\sqrt 51$ = 50 .
$y_1$ = denominator of the first convergent of the continued fraction fraction expansion of $\sqrt 51$ = 7
Now you can create infinitely many solutions as follows. The $n$-th solution is given by
$x_n$ = rational part of $(50 + 7 \sqrt 51)^n$
$y_n$ = irrational part of $(50 + 7 \sqrt 51)^n$
E.g. taking $n = 2$ gives the second solution pairs as $x_2 = 4999$, $y_2 = 700$.
A: From the continued fraction of $\sqrt{51}$
$$ \sqrt{51} = [7;\overline{7,14}] $$
and $[7;7]=\frac{50}{7}$ we get that all the integer solutions of $x^2-51y^2=1$ are associated with $x\pm y\sqrt{51} = (50\pm 7\sqrt{51})^m$. The minimal polynomial of $50+7\sqrt{51}$ over $\mathbb{Q}$ is given by $x^2-100x+1$, hence all the positive solutions of $x^2-51y^2=1$ can be generated through the recurrence
$$ x_0=1,y_0=0,x_1=50,y_1=7,\qquad (x_{n+2},y_{n+2}) = 100(x_{n+1},y_{n+1})-(x_n,y_n).$$
