Diophantine Equation

[flint topaz]

@woven oyster Look up Pell's equation... I have no idea how it works but it could be a possible way to answer this. However I don't think it's the correct method to use here because it seems pretty complicated.

[cunning girder]

I am not sure if it is correct approach or not but what I did is something like that:-
$x^2-5y^2=1\implies(x+y\sqrt5)(x-y\sqrt5)=1$
Which means,
if $x+y\sqrt5=1$ then $x-y\sqrt5=1$ too
solving further we get $x=1,y=0$
and if $x+y\sqrt5=-1$ then $x-y\sqrt5=-1$ too
solving further for this condition we get $x=-1,y=0$
Therefore, the set of solutions is $(x,y)\in(1,0),(-1,0)$

[fringe basin]

77²

[flint topaz]

I'm not sure this works because you dont find (x,y) = (9,4) which is a solution too... But nice work!

[cunning girder]

yeah that's the problem

[meager river]

Maybe you can substitute x² = u and y² = v getting u - 5v = 1 and then use Euclidean algorithm forwards and backwards

[cunning girder]

I think there could be infinitely many integral solutions to this

[woven oyster]

the pattern continues

[woven oyster]

ty

[cunning girder]

So there are really infinitely many solution?

[woven oyster]

yes

[cunning girder]

okok

[woven oyster]

(9,4) (161,72).....................

[blissful iron]

.close