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Could you please explain the proof of finding infinitely many solutions are existing for $a^2 - 10b^2 = 1$ or $4$ or $9$. Also, discuss, is there any relation between hyperbola and Pell's equation? If yes, how to find solution of $a$ and $b$ of $a^2 -10b^2 = 1$ by reducing into hyperbola form?

Thanks in advance.

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The main ingredient is to note that $$(a^2-10b^2)^n = (a+\sqrt{10}b)(a-\sqrt{10}b)^n = \left(A+\sqrt{10}B \right)\left(A-\sqrt{10}B \right) = A^2-10B^2$$ Once you have this, you can generate infinite solutions.


EDIT

Let us see how to get solutions for $a^2-10b^2 = 1 \tag{$\star$}.$ From guessing, we have $19^2-10 \cdot 6^2 = 1$, i.e., $(a,b) = (19,6)$. Square $(\star)$ to get, $$1^2 = (19 + 6\sqrt{10})^2(19 - 6\sqrt{10})^2 = \left( 721 + 228\sqrt{10}\right)\left( 721 - 228\sqrt{10}\right) = 721^2 - 10 \cdot 228^2$$ This gives us another solution as $(721,228)$. Keep proceeding like, i.e., instead of squaring, cube it, raise it to the fourth power and so on to get infinite solutions.


To get solutions for $a^2-10b^2 = 4 \tag{$\perp$},$ note that $a$ and $b$ have to be both even, i.e., let $a = 2a_1$ and $b=2b_1$ and find the infinite solutions to $a_1^2 - 10 b_1^2 = 1$ from above. Similarly, for $a^2-10b^2 = 9 \tag{$\dagger$},$ set $a = 3a_2$ and $b=3b_2$ and find infinite solutions to $a_2^2 - 10b_2^2 = 1$.

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  • $\begingroup$ ! left side of a^2 - 10b^2 = 1 can be written as (a+b) (a-b) form. what about right hand side 1 or 4 or 9. Who is the n? $\endgroup$ – Naroza_lary_sai May 11 '13 at 19:39
  • $\begingroup$ @Naroza_lary_sai Have updated the post and now it should be clear. $\endgroup$ – user17762 May 11 '13 at 19:50
  • $\begingroup$ ! wow! very interesting. Now, is there any relation between these equations and hyperbola equation. if there...please explain... $\endgroup$ – Naroza_lary_sai May 11 '13 at 19:53
  • $\begingroup$ @Naroza_lary_sai The canonical equation of a hyperbola is $$x^2 - c^2 y^2 = a^2$$ In your case, we have $a=1$ and $c={\sqrt{10}}$. $\endgroup$ – user17762 May 11 '13 at 19:55
  • $\begingroup$ There are integer solutions to $a^2-10b^2=9$ where $a, b$ are not divisible by $3$. In fact, we have $3$ fundamental solutions $(a, b)=(7, 2), (13, 4), (57, 18)$. $\endgroup$ – Ivan Loh May 12 '13 at 10:06

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