note
kvale
Good observation! It looks like Fermat's method (the method I
mentioned above) doesn't necessarily find all, or even any
of the factors. It is just a heuristic. On the other hand,
if you have an even number, you may extract factors of 2 until
you get an odd number, in which case (x+y) and (x-y) must
both be necessarily odd.
</p>
The generalization that gumby mentioned, x^2 = y^2 (mod p)
may have more sucess, but I don't know if it is exhaustive.
Here is code that implements it:
<code>
my $p = shift;
my %residues;
my %factors;
foreach my $x (2..$p) {
my $mod = ($x*$x) % $p;
if (exists $residues{$mod}) {
$factors{$x-$residues{$mod}}++ if $p % ($x-$residues{$mod}) == 0;
$factors{$x+$residues{$mod}}++ if $p % ($x+$residues{$mod}) == 0;
}
else {
$residues{$mod} = $x;
}
}
foreach (sort {$a <=> $b} keys %factors) {
print "$p = $_*", $p/$_, "\n";
}
</code>
For 54, this yields
<code>
1021% factor.pl 54
54 = 2*27
54 = 6*9
54 = 18*3
54 = 54*1
</code>
But this code is unsatisfying because it takes longer
than the simple trial by division. Thinking from a divide
and conquer point of view, it is probably faster to use
these methods to find a factorization, then recursively
find factorizations of the factors untill all are prime.
</p>
-Mark
173273
174047