Consider the quadratic Diophantine equation of the form:
x^2 – Dy^2 = 1
When D=13, the minimal solution in x is 649^2 – 13×180^2 = 1. It can be assumed that there are no solutions in positive integers when D is square.
Given a value of D, find the minimum value of X that gives a solution to the equation.
How exactly does, "6492 – 13×1802 = 1", as given in the example?
The 2s at the end of 649 and 180 used to be superscripts. I'm not quite sure when that changed, but it is fixed now. Thanks for the heads up on that.
No problem. Thanks for the fix!
Some tips. Continued fractions are the main way for finding the fundamental solutions to Pell's equations. And square roots have patterns in continued fractions.
I've used continued fractions and this link for my solution https://math.stackexchange.com/questions/2215918/continued-fraction-of-sqrt67-4/2216011#2216011
Sorry, I got stuck just trying to figure out how to use a switch statement again. This is not a solution.
Nice usage of DEAL and determination of intermediate integer solution to Pell's algorithm.
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