I'd like to find a recursive formula giving positive integer solutions to this Diophantine equation $$5L^2 - a^2 - 1 =0$$

It can be seen that I need $5L^2 - 1$ to be a square of a number $\in \mathbb N$.

The problem is, I never did this before, and I don't know where and how to begin.

I would be very much satisfied with a good read on finding recursive solutions to these type of equations, too.


There is an obvious solution $L=1$, $a=2$. You can expand $(2+\sqrt{5})^n$ as $a+L\sqrt{5}$ with odd $n$ to obtain more solutions. That is, from one solution $a+L\sqrt {5}$ you ontain the next as $$(a+L\sqrt{5})(2+\sqrt{5})^2=(a+L\sqrt{5})(9 +4\sqrt{5})\\=(9a+20L)+(4a+9L)\sqrt{5}.$$ Thus after $(2,1)$, you find $(38,17)$ and so on.

The trick behind this is that solutions are numbers in $\mathbb Z[\sqrt 5]$ with norm $-1$. Multiplying a solution with an element of norm $1$ produces a new solution.

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  • $\begingroup$ Fantastic, I should've noticed this; so is this how it is generally solved? $\endgroup$ – nullpotent Oct 20 '12 at 13:12
  • $\begingroup$ @iccthedral, pell equation. $\endgroup$ – sperners lemma Oct 24 '12 at 12:35

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