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 


USING: accessors arrays formatting kernel literals math
math.functions math.matrices math.ranges sequences ;
IN: rosettacode.pyth
 
CONSTANT: T1 { 
  {  1  2  2 }
  { -2 -1 -2 }
  {  2  2  3 }
}
CONSTANT: T2 {
  {  1  2  2 }
  {  2  1  2 }
  {  2  2  3 }
}
CONSTANT: T3 {
  { -1 -2 -2 }
  {  2  1  2 }
  {  2  2  3 }
}
 
CONSTANT: base { 3 4 5 }
 
TUPLE: triplets-count primitives total ;
: <0-triplets-count> ( -- a ) 0 0 \ triplets-count boa ;
: next-triplet ( triplet T -- triplet' ) [ 1array ] [ m. ] bi* first ;
: candidates-triplets ( seed -- candidates )
  ${ T1 T2 T3 } [ next-triplet ] with map ;
: add-triplets ( current-triples limit triplet -- stop )
  sum 2dup > [
   /i [ + ] curry change-total
   [ 1 + ] change-primitives drop t 
  ] [ 3drop f ] if ;
: all-triplets ( current-triples limit seed -- triplets )
  3dup add-triplets [ 
    candidates-triplets [ all-triplets ] with swapd reduce
  ] [ 2drop ] if ;
: count-triplets ( limit -- count )
  <0-triplets-count> swap base all-triplets ;
: pprint-triplet-count ( limit count -- )
  [ total>> ] [ primitives>> ] bi 
  "Up to %d: %d triples, %d primitives.\n" printf ;
: pyth ( -- )
  8 [1,b] [ 10^ dup count-triplets pprint-triplet-count ] each ;