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 


main :-
  garbage_collect(300_000_000),
  Data = [100, 1_000, 10_000, 100_000, 1_000_000, 10_000_000, 100_000_000],
  member(Max, Data),
  count_triples(Max, Total, Prim),
  printf("upto %d, there are %d Pythagorean triples (%d primitive.)%n", Max, Total, Prim),
  fail,
  nl.
 
count_triples(Max, Total, Prims) :-
  Ps = findall(S, (triple(Max, A, B, C), S is A + B + C)),
  Prims = Ps.len,
  Total = sum([Max div P : P in Ps]).
 
% - between_by/4
between_by(A, B, N, K) :-
  C = (B - A) div N,
  between(0, C, J),
  K = N*J + A.
 
% - Pythagorean triple generator
triple(P, A, B, C) :-
  Max = floor(sqrt(P/2)) - 1,
  between(0, Max, M),
  Start = (M /\ 1) + 1,
  Pm = M - 1,
  between_by(Start, Pm, 2, N),
  gcd(M, N) == 1,
  X = M*M - N*N,
  Y = 2*M*N,
  C = M*M + N*N,
  order2(X, Y, A, B),
  (A + B + C) =< P.
 
order2(A, B, A, B) :- A < B, !.
order2(A, B, B, A).