Question

Find a nontrivial integer solution to x^2 - 41y^2 = 1.

Answer 1

I know that an answer is x=2049, y=320, but is there a way to show this other than searching on a computer?

Answer 2

hmmm ... a hyperbola

Answer 3

what determines if a solution is non trivial?

Answer 4

a.k.a. x=1, y =0

Answer 5

Find the continued fraction for \[\sqrt{41}.\]You will get\[\sqrt{41} = 6 + \frac{1}{2 + \frac{1}{2 + \frac{1}{6+\sqrt{41}}}}\]which is a purely periodic continued fraction. Truncate the fraction at the first period, that is, we will use the approximation\[\sqrt{41} \approx 6 + \frac{1}{2+\frac{1}{2+0}} = \frac{32}{5}.\]This (x = 32, y = 5) will be our first solution, the fundamental solution. Notice that\[32^2 - 41 \cdot 5^2 = -1,\]which is similar, but not exactly what we are looking for. If you factor the equation like this\[(x - \sqrt{41}y)(x + \sqrt{41}y) = 1\]and think we are working in the ring\[\mathbb{Z}[\sqrt{41}]\]the problem becomes finding the units of that ring (that is, elements with norm 1, the norm for\[x \in \mathbb{Z}[\sqrt{41}]\]is defined as x times its conjugate, much like the norm of a complex number). If a number has norm 1, then if you take the nth power of that number it will still have norm 1, so if you find the fundamental\[(x_1, y_1)\] solution you can find any solution by using\[x_n + \sqrt{41}y_n = (x_1 + \sqrt{41}y_1)^n.\]You can find infinite solutions for the equation\[x^2 - 41 y^2 = \pm 1\]this way, and take the ones that satisfy\[x_n^2 - 41y_n^2 = 1\](for example, \[(32 + 5\sqrt{41})^2 = 2049 + 320\sqrt{41}\]and \[2049^2 - 41\cdot 320^2 = 1\]so\[x_2 = 2049, y_2 = 320\]which is the solution you already had).

Answer 6

Just out of curiosity, do you see a good way to do this using quadratic reciprocity?

Answer 7

I asked a professor and he said that it wasn't a quadratic reciprocity problem.