Diophantine equation of power 2 I want to solve the following equation: $x^2-17y^2=104$ where $x,y$ are integers. I don't know how to proceed here. I have tried with mod $13$ and mod $17$ but it doesn't work.
Thanks in advance!
 A: Above equation shown below:
$x^2-17y^2=104$
Above has parametric solution:
$x=[(11k^2-34k+187)/(k^2-17)]$
$y=[(k^2-22k-17)/(k^2-17)]$
For, $k=0$ we get:
$(x,y)=(11,1)$
A: COMMENT.- Modulo $17$ one has $x^2=2$ whose solutions in $\mathbb F_{17}$ are $x=6$ and $x=11$ so we have in $\mathbb Z$
$$x=17n+6\text{ and } x=17n+11$$ It follows respectively $$y^2=17n^2+12n-4\text{ and } y^2=17n^2+22n+1$$ Then we have the identities
$$(17n+6)^2-17(17n^2+12n-4)=104$$
$$(17m+11)^2-17(17m^2+22m+1)=104$$ It remains to study when $17n^2+12n-4$ and $17m^2+22m+1$ can be squares. 
Some solutions of the first, $17n^2+12n-4=y^2$ are $n=-5,-1,1,13$ which give respectively $y=19,1,5,55$ so $ x=\sqrt{17y^2+104}=79,11,23,227$. Thus one has, as examples, the four solutions $$(x,y)=(79,19),(11,1),(23,5),(227,55)$$ but there must be many more, infinitely many probably.
A: It is clear that $y=1$ and $x=11$ is a solution of $x^2-17y^2=104$.
As pointed out by others, so are $(x,y)=(23,5), (79,19),$ and $(227,55)$.
Once you have a solution $(x,y)$ to $x^2-17y^2=104$, another is $(33x+136y, 8x+33y)$.
That's because $33-17\times8^2=(33-8\sqrt{17})(33+8\sqrt{17})=1$, 
so if $x^2-17y^2=(x-y\sqrt{17})(x+y\sqrt{17})=104$ 
then $(33-8\sqrt{17})(x-y\sqrt{17})(33+8\sqrt{17})(x+y\sqrt{17})=104=$
$[(33x+8\times17y)-(8x+33y)\sqrt{17})][(33x+8\times17y)+(8x+33y)\sqrt{17}].$
The above information gives all of the solutions listed by Will Jagy.
