## Let $(a, b, c)$ 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?

### This is a somewhat complicated problem that requires knowledge of number theory, more specifically, Pythagorean triplets and Diophantine equations.

A Pythagorean triplet, if you’re not familiar with the term, is a set of three positive integers a, b and c that fits the rule a^2 + b^2 = c^2. In this case, your triangles are actually right triangles that fit this rule, which also means the hypotenuse (c) and the sum of two sides (a+b) determine a square.

The ‘hole’ you refer to is actually the difference between squares c^2 and (a+b)^2, which is 2ab.

Let’s rewrite the problem. You want to find the number of solutions for which (a,b,c) is a Pythagorean triplet, perimeter (a+b+c) less than 100,000,000 and 2ab is a perfect square.

This is actually an old problem of Diophantine equations. The solution is that the number of such triangles is a representation of the sum of all proper divisors of squares not exceeding 100,000,000 / 2.

In order to solve this, you’ll need to write a program to check the above condition for a, b, and c all less than 100,000,000. Then, count the number of solutions where 2ab is a perfect square.

Each unique Pythagorean triplet gives a solution to the problem. Therefore, the task is to count all of these Pythagorean triplets.

This is a complex problem that requires a significant amount of computation. As of today, there are no direct formulas to solve this problem. There is a “Generate and Test” process is considered appropriate.

Please note that this problem is not straightforward to solve and requires advanced mathematical techniques, hence, it may be hard to capture its details in a single response.

I would highly recommend referring to online resources that deal specifically with such Pythagorean triplets and Diophantine equation problems, as they might better illustrate the process and have code readily available.

##### More Answers:

Singleton DifferenceFibonacci Golden Nuggets

Special Isosceles Triangles