Not to be overly nitpicky, but this does not produce prime factors with any certainty. The first factor it provides is prime, the second only rarely is prime. So once you get the first prime factor, you have to run the second factor through this again until the second return value is 1.

Update: I somehow missed the part of the RSA challenge where they stated that the numbers they pose have only two prime factors, and for that, there indeed would be no need to calculate past one iteration.

And just so I'm not whining without contributing, here's a look at obtaining all the factors with the modulus routine. ;) 
#!/usr/bin/perl -w
use strict;

#note that this code is brutally slow

my $number = shift; #assume user knows to input a number

my @all_factors = factor( $number );
my $check_factor = 0;

while ($check_factor != 1 )
{
   $check_factor = pop @all_factors;
   push( @all_factors, factor( @check_factor ) );
}

if( @all_factors + 0 == 1 )
{
   print "$number is prime\n";
}
else
{
   print "$number: " .join( ',', @all_factors ). "\n";
}

exit;

sub factor
{ #one slight change to prevent an illegal modulus operation
    my $n; my $p;
    $n=$p=pop;
    return if( $n ==1 );
    1 while $n%--$p;
    $n/$p,$p;
}

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In reply to (ichimunki) Re x 2: RSA Factoring challenge by ichimunki
in thread RSA Factoring challenge by one4k4

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