Pell’s equation

Project Euler 139: Pythagorean tiles

Project Euler 139: Pythagorean tiles

In Project Euler There are loads of problems that end up with a number theoretic solution. Problem 139 is no exception to that.  The problem reads

Let (abc) represent the three sides of a right angle triangle with integral length sides. It is possible to place four such triangles together to form a square with length c.

For example, (3, 4, 5) triangles can be placed together to form a 5 by 5 square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5 square can be tiled with twenty-five 1 by 1 squares.

However, if (5, 12, 13) triangles were used then the hole would measure 7 by 7 and these could not be used to tile the 13 by 13 square.

Given that the perimeter of the right triangle is less than one-hundred million, how many Pythagorean triangles would allow such a tiling to take place?

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Posted by Kristian in Project Euler, 6 comments
Project Euler 138: Special isosceles triangles

Project Euler 138: Special isosceles triangles

Problem 138 of Project Euler reads

Consider the isosceles triangle with base length, b = 16, and legs, L = 17.

By using the Pythagorean theorem it can be seen that the height of the triangle, h = √(172 – 82) = 15, which is one less than the base length.

With b = 272 and L = 305, we get h = 273, which is one more than the base length, and this is the second smallest isosceles triangle with the property that h = b ± 1.

Find ∑ L for the twelve smallest isosceles triangles for which h = b ± 1 and b, L are positive integers.

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Posted by Kristian in Project Euler, 9 comments
Project Euler 94: Investigating almost equilateral triangles with integral sides and area.

Project Euler 94: Investigating almost equilateral triangles with integral sides and area.

I can realy feel that the problems become harder and harder. Problem 94 of Project Euler took me a good while to figure out. The problem is easy to understand and reads

It is easily proved that no equilateral triangle exists with integral length sides and integral area. However, the
5-5-6 has an area of 12 square units.

We shall define an almost equilateral triangle to be a triangle for which two sides are equal and the third differs by no more than one unit.

Find the sum of the perimeters of all almost equilateral triangles with integral side lengths and area and whose perimeters do not exceed one billion (1,000,000,000).

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Posted by Kristian in Project Euler, 25 comments
Project Euler 66: Investigate the Diophantine equation x^2 − Dy^2 = 1.

Project Euler 66: Investigate the Diophantine equation x^2 − Dy^2 = 1.

Based on the problem description for Problem 66 of Project Euler I thought we had left the continued fractions for a while. Once I started reading up on the maths behind it and trying to solve the problem I got quite a lot wiser. But let’s start from the beginning and look at the problem description which reads

Consider quadratic Diophantine equations of the form:

x2 – Dy2 = 1

For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.

It can be assumed that there are no solutions in positive integers when D is square.

By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:

32 – 2×22 = 1
22 – 3×12 = 1
92 – 5×42 = 1
52 – 6×22 = 1
82 – 7×32 = 1

Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.

Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.

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Posted by Kristian in Project Euler, 12 comments