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 


var sc = System.clock
var total = 0
var prim = 0
var maxPeri = 0
 
var newTri // recursive function so needs to be declared before it can be called
newTri = Fn.new { |s0, s1, s2|
    var p = s0 + s1 + s2
    if (p <= maxPeri) {
        prim = prim + 1
        total = total + (maxPeri/p).floor
        newTri.call( 1*s0-2*s1+2*s2,  2*s0-1*s1+2*s2,  2*s0-2*s1+3*s2)
        newTri.call( 1*s0+2*s1+2*s2,  2*s0+1*s1+2*s2,  2*s0+2*s1+3*s2)
        newTri.call(-1*s0+2*s1+2*s2, -2*s0+1*s1+2*s2, -2*s0+2*s1+3*s2)
    }
}
 
maxPeri = 100
while (maxPeri <= 1e10) {
    prim = 0
    total = 0
    newTri.call(3, 4, 5)
    var secs = (System.clock - sc).round
    System.print("Up to %(maxPeri):  %(total) triples, %(prim) primitives, %(secs) seconds")
    maxPeri = 10 * maxPeri
}