I have a question that I have problem with in number theory about Diophantine,and Pell's equations. Any help is appreciated!

We suppose $n$ is a fixed non-zero integer, and suppose that $x^2_0 - 3 y^2_0 = n$, where $x_0$ and $y_0$ are bigger than or equal to zero. Let $x_1 = 2 x_0 + 3 y_0$ and $y_1 = x_0 + 2 y_0$. We need to show that we have $x^2_1 - 3 y^2_1 = n$, with $x_1>x_0$, and $y_1>y_0$. Also, we need to show then that given $n$, the equation $x^2 - 3 y^2 = n$ has either no solutions or infinitely many solutions. Thank you very much!

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    $\begingroup$ Just out of curiosity: How long did you spend trying to do this problem on your own before posting? $\endgroup$ – Arturo Magidin Apr 21 '11 at 20:02
  • $\begingroup$ You should substitute $x=2x_0+3y_0$, $y=x_0+2y_0$ in the expression $x^2-3y^2$, simplify, see what happens. $\endgroup$ – André Nicolas Apr 21 '11 at 20:02
  • $\begingroup$ @Arturo: I did try but I think I did mistake somewhere because I couldn't simplify the equation. Thanks! $\endgroup$ – kira Apr 21 '11 at 20:11
  • $\begingroup$ @user6312:I did the same thing but got a problem. I'll try later again. Thanks! $\endgroup$ – kira Apr 21 '11 at 20:11
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    $\begingroup$ Next time, please say what you tried and why things are not working out. Here, you could easily have posted your attempt, and people could have pointed out if (or where) there was a mistake. You'd learn a lot more that way. $\endgroup$ – Arturo Magidin Apr 21 '11 at 20:12

The fact that if $x_1=2x_0+3y_0$ then $x_1\gt x_0$ is immediate: you cannot have both $x_0$ and $y_0$ zero; likewise with $y_1$.

That $x_1^2+3y_1^2$ is also equal to $n$ if you assume that $x_0^2 - 3y_0^2=n$ should follow by simply plugging in the definitions of $x_1$ and $y_1$ (in terms of $x_0$ and $y_0$), and chugging.

Finally, what you have just done is show that if you have one solution, you can come up with another solution. Do you see how this implies the final thing you "need to show"?

  • $\begingroup$ How to show it has no solutions or infinetely many solutions? Thanks! $\endgroup$ – kira Apr 24 '11 at 23:15
  • $\begingroup$ @kira: The entire process tells you how to go from one solution to another. Keep going. If you have at least one solution, how many different solutions will you have? $\endgroup$ – Arturo Magidin Apr 25 '11 at 2:00
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    $\begingroup$ If we have a solution, then we can find another one with $x_1>x_0$, and $y_1>y_0$ in the same quadrant. Thus, we have infinitely many. Is anything to be added since we can find a new solution every time with bigger $x_i$ and $y_i$? $\endgroup$ – user9636 Apr 25 '11 at 4:11

HINT $\: $ Put $\rm\: z = x+\sqrt{3}\ y\:,\:$ norm $\rm\:N(z)\: = z\:z' = x^2 - 3\ y^2\:.\:$ Then $\rm u = 2 + \sqrt{3}\ \Rightarrow\ N(u) = u\:u' = 1\:$ so $\rm\ N(u\:z)\ =\ (u\:z)\:(u\:z)' =\ u\:u'\:z\:z'\ =\ z\:z'\:,\:$ where $\rm\ u\:z\ =\ 2\:x+3\:y + (x+2\:y)\ \sqrt{3}\:.\:$ Therefore the composition law (symmetry) $\rm\ z\to u\:z\ $ on the solution space $\rm\:\{z\ :\ N(z) = n\}$ arises simply by multiplying by an element of $\rm\:u\:$ of norm $1\:,\:$ using the multiplicativity of the norm: $$\rm\ N(u) = 1\ \ \Rightarrow\ \ N(u\:z)\ =\ N(u)\:N(z)\ =\ N(z) = n$$


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