prove that $x^2-2y^2=7$ has infinitely many solutions in integers.
$a$ and $b$ are two non-zero rational numbers.Then if the equation $ax^2+by^2=0$ has a nonzero rational solution $(x_0,y_0)$ then for any rational $t$, $ax^2+by^2=t$ also has a non zero rational solution $(x_t,y_t)$.Prove this.
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$\begingroup$ $K = \mathbb{Q}(\sqrt{2})$. What properties of $K$ and the field norm $N_{K/\mathbb{Q}}$ do you know ? $\endgroup$– reunsCommented Oct 10, 2017 at 19:43
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$\begingroup$ It isn't clear what part of this is your question. What are you stuck on? $\endgroup$– ThéophileCommented Oct 10, 2017 at 19:43
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$\begingroup$ Unfortunately, I feel I had to flag this. As you know, this isn’t a site that will solve your homework for you, and some effort or examples of your own thoughts must be show, as well as an explanation of context. For more information, please read how to ask a good question. At any rate, I wish the best of luck to you, and please continue to contribute to our wonderful site! $\endgroup$– gen-ℤ ready to perishCommented Oct 10, 2017 at 20:04
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2 Answers
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The first equation is what is generally known as the Pell's equation. For this it is easy to see that $(3,1)$ is a solution. Then one can easily check that any $(x_n, y_n)$ satisfying $$x_n+\sqrt{2}y_n=(3+\sqrt{2})(3+2\sqrt{2})^n$$ will also be solutions to the given equation, hence infinitely many integer solutions.
For the second question, please share your thoughts so that it will be easier to see where you need help.
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$\begingroup$ your answer will work if I replace 7 by 1 $\endgroup$– ShBhCommented Oct 10, 2017 at 19:49
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$\begingroup$ @reuns thanks for pointing the error. I will fix it. $\endgroup$– Anurag ACommented Oct 10, 2017 at 19:54
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$\begingroup$ @ShubhrajitBhattachrya I have fixed the error. $\endgroup$– Anurag ACommented Oct 10, 2017 at 19:56
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Define the sequences $(x_n,y_n)$ recursively by ($x_1=3$ and $y_1=1$) \begin{eqnarray*} x_{n+1}=3x_n+4y_n \\ y_{n+1}=2x_n +3y_n. \end{eqnarray*} It easy to show by induction that the pairs $(x_n,y_n)$ will satisfy $x^2-2y^2=7$. \begin{eqnarray*} x_{n+1}^2-2y_{n+1}^2=(3x_n+4y_n)^2-2( 2x_n +3y_n)^2 \\ =9x_{n}^2+24x_ny_n+16y_{n}^2-2(4x_{n}^2+12x_ny_n+9y_{n}^2)=x_{n}^2-2y_{n}^2=7. \end{eqnarray*}
answered Oct 10, 2017 at 20:08