Problem 1215. Diophantine Equations (Inspired by Project Euler, problem 66)

Created by James

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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.

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Last Solution submitted on Jun 22, 2020

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Informaton on 3 May 2017

How exactly does, "6492 – 13×1802 = 1", as given in the example?

James on 8 May 2017

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.

Informaton on 18 Aug 2017

No problem. Thanks for the fix!

Rafael Siqueira Telles Vieira on 22 Jun 2020 at 6:21

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.

Solution Comments

Solution 2594281

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Rafael Siqueira Telles Vieira on 22 Jun 2020 at 5:45

I've used continued fractions and this link for my solution https://math.stackexchange.com/questions/2215918/continued-fraction-of-sqrt67-4/2216011#2216011

Solution 1178675

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1 Comment

Informaton on 3 May 2017

Sorry, I got stuck just trying to figure out how to use a switch statement again. This is not a solution.

Solution 193603

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1 Comment

Richard Zapor on 20 Jan 2013

Nice usage of DEAL and determination of intermediate integer solution to Pell's algorithm.

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Problem 1215. Diophantine Equations (Inspired by Project Euler, problem 66)

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Problem 1215. Diophantine Equations (Inspired by Project Euler, problem 66)

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Last 200 Solutions

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Problem Recent Solvers43

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