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I want to show that the formula

$$ | a^2 - 26\cdot b^2| = 1$$

has infinitely many solutions $(a, b) \in \mathbb{Z}^2$.

First I tried to solve the formula for one of the two variables, to get something like $a = ±\sqrt{±1 + 26 b^2}$ or $b = \sqrt{-\frac{±1 - a^2}{26}} $, but both doesn't look very promising, seeing as I would now need to prove that there are infinitely many integer solutions for $b$ (or $a$) so that the respective other one is also an integer.

I did, however, get some solutions. I could (by more or less trying out values) find out that $(5, 1), (51, 10)$ are two solutions. (And therefore also $(-5, 1)$ and so on.) So maybe there is a recursive way to get new values, based on the ones I already have, i.e. a formula $\phi(a, b)$ so that, when $(a, b)$ is an integer solution, $\phi(a, b)$ is aswell? (We would probably need an extra condition like that the components of $\phi(a, b)$ are strictly larger than $a, b$, so that we really get new solutions and not the ones we already had before.)

Seeing as I don't actually want to find a formula or so for the actual solutions, it would also be sufficient if there's a proof that there are infinitely many solutions that doesn't involve actually finding any of the solutions. But so far, I haven't had any luck with that.

(I wasn't entirely sure what tags to choose; the problem seemed analytical to me if other tags are seen as more suiting, please feel free to correct them.)

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$(5a+26b)^2-26(a+5b)^2=-(a^2-26b^2)$. So you can use a solution to generate another larger solution. Eg starting with $(5,1)$, you get $(51,10)$.

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  • $\begingroup$ @mathlove Thanks. Fixed. $\endgroup$ – almagest Jun 11 '16 at 10:38
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$|a^2-26b^2|=1$ has infinitely many solutions if and only if $a^2-26b^2=1$ or $a^2-26b^2=-1$ has infinitely many solutions.

The latter are known as the Pell equations (Wikipedia link) and have been stuided thoroughly. The Wikipedia page has full explanations on how to find parametrizations. You might also want to look into the mathworld site.

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If $\alpha=a+b\sqrt{26}$, let $N(\alpha)=a^2 - 26b^2$. Then $N(\alpha\beta)=N(\alpha)N(\beta)$.

Now, by inspection, if $\theta = 5+\sqrt{26}$,then $N(5+\sqrt{26})=-1$.

Therefore, $|N(\theta^n)|=|N(\theta)^n|=1$ and so $\theta^n$ gives you an infinitude of solutions.

Note that $\theta^n=a_n + b_n \sqrt{26}$, with $a_n,b_n \in \mathbb Z$. In fact, $$ \begin{array}% a_{n+1} =& 5a_n+26b_n, && a_0 =& 1 \\ b_{n+1} =& a_n + 5b_n, && b_0 =& 0 \end{array} $$

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    $\begingroup$ right. and Cayley-Hamilton for your two by two matrix gives separate linear recursions, one for your $a_n$ and one for $b_n$ $\endgroup$ – Will Jagy Jun 11 '16 at 18:52
  • $\begingroup$ which you may well have known, I'm just sayin' $\endgroup$ – Will Jagy Jun 11 '16 at 18:58
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    $\begingroup$ @WillJagy, yes, good point! $\endgroup$ – lhf Jun 11 '16 at 19:01
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Here is what you did not want. All of them. $$ x_n^2 - 26 y_n^2 = (-1)^n, $$ with $$ x_0 = 1, \; \; x_1 = 5, \; \; x_2 = 51, \; \; x_3 = 515, \; \; x_4 = 5201, \ldots $$ $$ y_0 = 0, \; \; y_1 = 1, \; \; y_2 = 10, \; \; y_3 = 101, \; \; y_4 = 1020, \ldots $$ with $$ x_{n+2} = 10 x_{n+1} + x_n, $$ $$ y_{n+2} = 10 y_{n+1} + y_n. $$

To arrive full circle, the pair of recursions are simply saying that $x_n/y_n$ is a convergent for $\sqrt {26},$ and the "digits" for that continued fraction are $$ \left[5; 10, 10, 10, 10, \ldots \right] $$

Alright, curious what happens in general. If we have positive $N$ such that negative Pell is solvable, and we take the smallest solution $$ u^2 - N v^2 = -1 $$ with positive $u,v$ then we have $x_n - N y_n^2 = (-1)^n,$ where $$ x_0 = 1, \; \; x_1 = u, \; \; x_2 = 2u^2 + 1, \; \; x_3 = 4 u^3 + 3u, \; \; x_4 = 8u^4 + 8u^2 + 1 ,\ldots $$ $$ y_0 = 0, \; \; y_1 = v, \; \; y_2 = 2uv, \; \; y_3 = 4 u^2 v + v, \; \; y_4 = 8 u^3 v + 4 u v, \ldots $$ and $$ x_{n+2} = 2u x_{n+1} + x_n, $$ $$ y_{n+2} = 2u y_{n+1} + y_n. $$

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