Thanks for the tip on $b=$a... although I was looking for all possible permutations (ok to switch values for $a and $b), it helps to only grab unique. Guess I could toss an ARGV in for "uniqueness".

As far as the square root, you could also use the built-in sqrt() function. Although yes, I'm still trying to alleviate the float problem (without using modules). ;-)

-fuzzyping

UPDATE:

Just realized my original logic with the $c loop was greatly flawed as well... it's limited by $max. For example, where $max=10, it should find:

3 4 5 is a winner
6 8 10 is a winner

But it was only finding /3 4 5/ due to the limit on $c (duh). Removing the $c loop has helped, but required me to cheat on the final confirmation (I didn't want to use any modules). Using sqrt and then avoiding any floats is the answer, but requires that I test $c as a string. :-(

#!/usr/bin/perl

use strict;

my ($a, $b, $c);
my $max = shift || die "Usage: pythag.pl <max len>\n";

for ($a=1; $a<$max; $a++) {
        for ($b=$a; $b<$max; $b++) {
                $c = sqrt(($a**2) + ($b**2));
                unless ($c =~ /\./) {
                        print "$a $b $c is a winner\n";
                }
        }
}
[download]

The final result is only one line smaller, but saves .243ms per successful match.

-fuzzyping

UPDATE:

Thanks to tye and Sinister (don't know how I missed it before), I've implemented the int() check instead (keeping it within the numeric handling constraints I prefer)...

#!/usr/bin/perl

use strict;

my ($a, $b, $c);
my $max = shift || die "Usage: pythag.pl <max len>\n";

for ($a=1; $a<$max; $a++) {
        for ($b=$a; $b<$max; $b++) {
                $c = sqrt(($a**2) + ($b**2));
                if (int($c) == $c) {
                        print "$a $b $c is a winner\n";
                }
        }
}
[download]

Thanks to all!

-fuzzyping