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 


 
# syntax: GAWK -f PYTHAGOREAN_TRIPLES.AWK
# converted from Go
BEGIN {
    printf("%5s %11s %11s %11s %s\n","limit","limit","triples","primitives","seconds")
    for (max_peri=10; max_peri<=1E9; max_peri*=10) {
      t = systime()
      prim = 0
      total = 0
      new_tri(3,4,5)
      printf("10^%-2d %11d %11d %11d %d\n",++n,max_peri,total,prim,systime()-t)
    }
    exit(0)
}
function new_tri(s0,s1,s2,  p) {
    p = s0 + s1 + s2
    if (p <= max_peri) {
      prim++
      total += int(max_peri / p)
      new_tri(+1*s0-2*s1+2*s2,+2*s0-1*s1+2*s2,+2*s0-2*s1+3*s2)
      new_tri(+1*s0+2*s1+2*s2,+2*s0+1*s1+2*s2,+2*s0+2*s1+3*s2)
      new_tri(-1*s0+2*s1+2*s2,-2*s0+1*s1+2*s2,-2*s0+2*s1+3*s2)
    }
}
 