# Problem 218: Perfect right-angled triangles

Consider the right angled triangle with sides a=7, b=24 and c=25. The area of this triangle is 84, which is divisible by the perfect numbers 6 and 28.
Moreover it is a primitive right angled triangle as gcd(a,b)=1 and gcd(b,c)=1.
Also c is a perfect square.

We will call a right angled triangle perfect if

• it is a primitive right angled triangle
• its hypotenuse is a perfect square
We will call a right angled triangle super-perfect if
• it is a perfect right angled triangle and
• its area is a multiple of the perfect numbers 6 and 28.
How many perfect right-angled triangles with c <= 10^16 exist that are not super-perfect?

# My Algorithm

I wrote a small program to check all such triangle with c <= 10^8 and found zero perfect right-angled triangles that are not super-perfect.
See below for the simple code, most of it was copied from problem 86.

Even trying some randomized values for m and n failed to find such a triangle.

Then I went back to my highly valued "paper and pencil" technique and found a relationship:
each primitive triple is defined by
(1) a = m^2 - n^2
(2) b = 2mn
(3) c = m^2 + n^2
(4) gcd(m,n) = 1
(5) (m mod 2) != (n mod 2)
Those equations were already part of multiple Project Euler problems, such as problem 86.

This time there are more restrictions:

• c is a perfect square, so there is an integer d such that c = d^2
• the area of each right-angled triangle is A = ab / 2
• A mod 6 == 0 and A mod 28 == 0
The last restriction can be simplified: the least common multiple lcm(6,28) = 6 * 28 / gcd(6,28) = 84, therefore A mod 84 == 0 and
the area A must be a multiple of 84:
(6) A == 0 mod 84
(7) ab/2 == 0 mod 84
(8) ab == 0 mod 168

I have to find all perfect right-angled triangle where the area is a multiple of 168 (and the hypotenuse < 10^16)

Since c is a perfect square with c = d^2 I can rewrite c = m^2 + n^2 as
(9) d^2 = m^2 + n^2

This is actually a primitive triple again - because of gcd(m, n) = 1. So can repeat the procedure again and there must be some x and y such that
(10) m = x^2 - y^2
(11) n = 2xy
(12) d = x^2 + y^2

Substituting m and n in (1) and (2):
(13) a = (x^2 - y^2)^2 - (2xy)^2
(14) b = 2 * (x^2 - y^2) * 2xy

Therefore the area of these triangles becomes (see (8) ):
(15) ((x^2 - y^2)^2 - (2xy)^2) * 2 * (x^2 - y^2) * 2xy == 0 mod 168
(16) ((x^2 - y^2)^2 - (2xy)^2) * (x^2 - y^2) * xy == 0 mod 42

I wrote two nested loops iterating over all 42^2 basic pairs (x mod 42, y mod 42) (see countNotMod42) - and actually of them produce zero in equation (16).
That means that there are no solutions, no matter whether the limit is 10^8, 10^16 or infinity.

## Alternative Approaches

Nayuki's proof is almost identical to my concept. Some of my ideas are solved by a program (multiple of 42) whereas he showed the same in a mathematical way.
It's also important to note that he found his proof a few years earlier.
I was kind of surprised when my program returned zero and even suspected a bug because I didn't fully trust countNotMod42.

## Note

countNotMod42 fails when x or y are int (or unsigned) instead of long long because of overflows. It cost me half an hour to realize that problem.
This function also contains a constant named Multiplier which can be ignored → set it to 1.

# Interactive test

This feature is not available for the current problem.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       #include <iostream>
#include <cmath>

// return number of pairs (x,y) such that equation (16) is not zero mod 42
unsigned int countNotMod42()
{
unsigned int result = 0;

const auto Multiplier = 1; // no more than 5 because then long long overflows
for (long long x = 0; x < 42 * Multiplier; x++)
for (long long y = 0; y < 42 * Multiplier; y++)
{
auto zero = ((x*x - y*y)*(x*x - y*y) - 2*x*y*2*x*y) * (x*x - y*y) * x * y;
if (zero % 42 != 0)
result++;
}
return result;
}

// greatest common divisor
template <typename T>
T gcd(T x, T y)
{
while (x != 0)
{
auto temp = x;
x = y % x;
y = temp;
}
return y;
}

int main()
{
// count how many possible pairs (x,y) could generate (a,b,c) according to equation (16)
std::cout << countNotMod42() << std::endl;
return 0;

//#define SEARCH
#ifdef  SEARCH
// ---------- try to find a non super-perfect triangle up to 10^8 ---------
// note: code currently not reached because the program exited three lines ago
unsigned long long limit = 100000000;
std::cin >> limit;

unsigned long long countNotSuperPerfect = 0;

// find basic Pythagorean triples (code copied from problem 86)
for (unsigned long long m = 1; m <= sqrt(2*limit); m++)
for (unsigned long long n = 1; n < m; n++)
{
if (m % 2 == n % 2)
continue;
if (gcd(m, n) != 1)
continue;

// two sides
auto a = m*m - n*n;
auto b = 2*m*n;
auto c = m*m + n*n;
if (c > limit)
break;

// is c a perfect square ?
unsigned long long cRoot = sqrt(c);
if (cRoot * cRoot != c)
continue;

// is area a multiple of 6 and 28 ?
auto area = a * b / 2;
if (area % 6 != 0 || area % 28 != 0) // can be combined: area % lcm(6, 28) = area % 84
{
countNotSuperPerfect++;
continue;
}
}

// show result
std::cout << countNotSuperPerfect << std::endl;
return 0;
#endif
}


This solution contains 11 empty lines, 11 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

August 28, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 55% (out of 100%).

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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

more about me can be found on my homepage, especially in my coding blog.
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