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 $4$ )
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$\begingroup$ What is a "powerful number"? $\endgroup$ – GFauxPas May 14 '15 at 15:55
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$\begingroup$ @GFauxPas : en.wikipedia.org/wiki/Powerful_number $\endgroup$ – user228168 May 14 '15 at 15:56
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$\begingroup$ Try looking up Pell's equation. $\endgroup$ – Brandon Thomas Van Over May 14 '15 at 16:07
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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?
answered May 14 '15 at 16:05
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$\begingroup$ 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 ? $\endgroup$ – user228168 May 14 '15 at 16:16
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$\begingroup$ Iterating the transformation mod $3$ I get $(1,1) \to (0,2) \to (2,1) \to (2,2) \to (0,1) \to \ldots$. $\endgroup$ – Robert Israel May 14 '15 at 19:36
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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.
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1$\begingroup$ where did those $485 , 132 , 1782 $ come from ? $\endgroup$ – user228168 May 15 '15 at 4:51
