A Pythagorean triple is defined as three positive integers 



(
a
,
b
,
c
)


{\displaystyle (a,b,c)}

 where 



a
<
b
<
c


{\displaystyle a<b<c}

, and 




a

2


+

b

2


=

c

2


.


{\displaystyle a^{2}+b^{2}=c^{2}.}



They are called primitive triples if 



a
,
b
,
c


{\displaystyle a,b,c}

 are co-prime, that is, if their pairwise greatest common divisors 





g
c
d


(
a
,
b
)
=


g
c
d


(
a
,
c
)
=


g
c
d


(
b
,
c
)
=
1


{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}

.

Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (





g
c
d


(
a
,
b
)
=
1


{\displaystyle {\rm {gcd}}(a,b)=1}

).  

Each triple forms the length of the sides of a right triangle, whose perimeter is 



P
=
a
+
b
+
c


{\displaystyle P=a+b+c}

.



Task

The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.



Extra credit
 
Deal with large values.   Can your program handle a maximum perimeter of 1,000,000?   What about 10,000,000?   100,000,000?

Note: the extra credit is not for you to demonstrate how fast your language is compared to others;   you need a proper algorithm to solve them in a timely manner.



Related tasks

   Euler's sum of powers conjecture 
   List comprehensions
   Pythagorean quadruples 


#include <stdio.h>
#include <stdlib.h>
 
typedef unsigned long long xint;
typedef unsigned long ulong;
 
inline ulong gcd(ulong m, ulong n)
{
    ulong t;
    while (n) { t = n; n = m % n; m = t; }
    return m;
}
 
int main()
{
    ulong a, b, c, pytha = 0, prim = 0, max_p = 100;
    xint aa, bb, cc;
 
    for (a = 1; a <= max_p / 3; a++) {
        aa = (xint)a * a;
        printf("a = %lu\r", a); /* show that we are working */
        fflush(stdout);
 
        /*  max_p/2: valid limit, because one side of triangle
         *  must be less than the sum of the other two
         */
        for (b = a + 1; b < max_p/2; b++) {
            bb = (xint)b * b;
            for (c = b + 1; c < max_p/2; c++) {
                cc = (xint)c * c;
                if (aa + bb < cc) break;
                if (a + b + c > max_p) break;
 
                if (aa + bb == cc) {
                    pytha++;
                    if (gcd(a, b) == 1) prim++;
                }
            }
        }
    }
 
    printf("Up to %lu, there are %lu triples, of which %lu are primitive\n",
        max_p, pytha, prim);
 
    return 0;
}