my($a,$b,@r,$c)=@_;for my$i(0..$#$a){$$c[$i+$_]+=$$a[$i]*$$b[$_]for 
+0..$#$b}@r?p($c,@r):$c
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   MeowChow                                   
               s aamecha.s a..a\u$&owag.print

(Posted at MeowChow's request.)

My best is 77:

sub p{
@m=1;for$p(@_){my@t;for$i(0..@m){my$j;$t[$i+$j++]+=$_*$m[$i]for@$p}@m=
+@t}[@m]
}
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Note that this introduces 0's through a fencepost error, but they don't change which polynomial is represented. I think this is fair, but if you think that is cheating, you can not save that character:

sub p{
@m=1;for$p(@_){my@t;for$i(0..$#m){my$j;$t[$i+$j++]+=$_*$m[$i]for@$p}@m
+=@t}[@m]
}
[download]

The trick lies in finding ways to not work through explicit lookups by index, and in finding ways to not access arrays through references. In fact there is not a single lookup by index of an element in an array reference. (It was cheaper to create and manually increment the index variable.)

BTW note that the statement of the rules anticipated and forbade saving a character by ending with \@m without making @m a private variable.

Finally at a request from chatter, here is the solution broken out and commented:

sub p{
  @m=1;                       # Start the product at 1.
  for$p(@_){                  # Loop over the polynomials.
    my@t;                     # Create a private temp array.
    for$i(0..@m){             # Loop over the indexes of @m.
      my$j;                   # Create the *other* index var.
      $t[$i+$j++]+=           # Manually increment $j while..
                              # Adding to the index of @t..
        $_*$m[$i]             # 2 terms multiplied together..
          for@$p              # for all the terms in the..
    }                         # other polynomial.
    @m=@t                     # Make the temp array our new
  }                           # product.
  [@m]                        # Return our answer in the
}                             # desired form.
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UPDATE
Never say you are done, 2 more characters:

sub p{
@m=1;for$p(@_){my@t;for$i(0..@m){$j=$i;$t[$j++]+=$_*$m[$i]for@$p}@m=@t
+}[@m]
}
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UPDATE 2
(This is a couple of days later.) Truly never say never, there were 2 more wasted characters to 73:

sub p{
@m=1;for$p(@_){$i=my@t;for$,(@m){$j=$i++;$t[$j++]+=$_*$,for@$p}@m=@t}[
+@m]
}
[download]

D'oh, I see where I was going wrong, I was reading it as [3,2] being 3x+2. Now I see ability to whack zeros with no problem :D

However, to nitpick, the 75 char solution (update above), isn't strict; you need to add 12 characters to strict-ify it. (my@m, for my $i, for my $p, my$j ), so to compare with the other solutions, at least mine being all strict, you're at 87.

update-fixed bad square brackets.

---

Dr. Michael K. Neylon - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain

By "return a polynomial" I mean return one in the given representation, ie it should be an anonymous array. So you need to change the last @c for \@c. (Makes this 110 characters.) The rule about different polynomials means that if you call this twice, the second answer should be a reference to a different array than the first.

sub p {
  my($a,$b)=(shift,shift);
  my@c=map{my($d,$e);for$e(0..$_){$d+=$$a[$e]*$$b[$_-$e]}$d}0..$#$a+$#
+$b;
  @_?p(\@c,@_):\@c
}
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Simple improvement:

my($a,$b)=(shift,shift);
my$a=shift;my$b=shift; # a bit shorter
my$a=pop;my$b=pop; # even shorter since multiplaction is commutative
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        - tye (but my friends call me "Tye")