Problem Statement

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}

.

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.

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.

Let's start with the solution:

sub gcd {
    my ($n, $m) = @_;
    while($n){
        my $t = $n;
        $n = $m % $n;
        $m = $t;
    }
    return $m;
}

sub tripel {
    my $pmax  = shift;
    my $prim  = 0;
    my $count = 0;
    my $nmax = sqrt($pmax)/2;
    for( my $n=1; $n<=$nmax; $n++ ) {
        for( my $m=$n+1; (my $p = 2*$m*($m+$n)) <= $pmax; $m+=2 ) {
            next unless 1==gcd($m,$n);
            $prim++;
            $count += int $pmax/$p;
        }
    }
    printf "Max. perimeter: %d, Total: %d, Primitive: %d\n", $pmax, $count, $prim;
}

tripel 10**$_ for 1..8;