# Does the equation $3x^2-2y^2=1$ has infinitely many integer solutions such that $3|x$ ?

How to show that the equation $3x^2-2y^2=1$ has infinitely many integer solutions such that $3|x$ ? ( If this can be shown then solutions of $12x^2-8y^2=4$ give infinitely many powerful numbers differing by $4$ )

If $(x,y)$ is a solution, then so is $(5x+4y, 6x+5y)$. Starting with $(1,1)$, you get a sequence of solutions. What is $x$ mod $3$ for these?
• the $x$'s are not coming to be $0$ mod $3$ , if I assume $3|5x+4y$ , then $5(5x+4y)+4(6x+5y) \equiv x+y$ (mod $3$ ) .... what should I do then ? – user228168 May 14 '15 at 16:16
• Iterating the transformation mod $3$ I get $(1,1) \to (0,2) \to (2,1) \to (2,2) \to (0,1) \to \ldots$. – Robert Israel May 14 '15 at 19:36
Put another way, $$27 \cdot 3^2 - 2 \cdot 11^2 = 1.$$ Given one $(x,y)$ pair with $$27 x^2 - 2 y^2 = k,$$ we get the next pair, of infinitely many, from $$(485x + 132 y, 1782x+485 y).$$
With your $k=1$ you need start with the single seed pair $(3,11)$ to get all $(x,y)$ pairs with $x,y>0.$
For composite $k,$ you may need two or more seeds, but a small finite number in any case.
• where did those $485 , 132 , 1782$ come from ? – user228168 May 15 '15 at 4:51