Problem 218: Perfect right-angled triangles

(see projecteuler.net/problem=218)

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

We will call a right angled triangle super-perfect if
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:

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.

#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
August 28, 2017 added comments

Difficulty

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

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