Ahh! I have deciphered the meaning of your koan. I get a compilation error when I take the "our" out of the middle of the recursive regex:

$ vi p6.pl
"p6.pl" 36 lines, 925 characters 
#!/opt/perl5.16/bin/perl

use strict;
use warnings;

our $paren = qr/                # Need declared variable with use stri
+ct.
      \(
        (
           [^()]+               # Not parens
         |
           (??{ $paren })       # Another balanced group (not interpol
+ated yet)
        )*
      \)
    /x;                         # 'x' means ignore whitespace, comment
+s.


my $stuff = "On the outside now then (we go( in( and in (&stop)(awhile
+) ( furthe
r ))) but still (here) ) and now (for a while) we are out again.";
$stuff =~ /($paren)[^()]*($paren)/;

print "----------\n";
print "$stuff\n";
print "----------\n";
"p6.pl" 27 lines, 650 characters 
$ ./p6.pl 
Variable "$paren" is not imported at (re_eval 1) line 2.
Global symbol "$paren" requires explicit package name at (re_eval 1) l
+ine 2.
Compilation failed in regexp at ./p6.pl line 14.
$
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If I delete the "use strict;" the program compiles/runs, but still does not match the 2nd regex. I also tried making a duplicate of $paren called $par2, and used it instead of $paren to set $2, and that doesn't work either, $2 is still blank.

... $2 is still blank.

You have to properly account for all capturing groups in the overall regex, properly counting capture groups in any interpolated regex. If you have a capturing group within the interpolated recursive regex (not, IMHO, necessary), then you want to access the 3rd capture group variable. However, a capture group in the recursive regex screws up the more general
    @p = $s =~ m{ $r4 }xmsg;
extraction regex. See examples below.

>perl -wMstrict -le
"my $s = 'x(y) (a(b)) ()() q (a(b)c()(d(e(f)g))h) q';
 ;;
 our $r3 = qr{
   \(
     (?:
       [^()]+
       |
       (??{ our $r3 })
     )*
   \)
   }xms;
 ;;
 my @p = $s =~ m{ $r3 }xmsg;
 print qq{'$_'} for @p;
 print '--------';
 ;;
 $s =~ m{ ($r3) [^()]* ($r3) }xms;
 print qq{1 '$1'   2 '$2'};
 print '--------';
 ;;
 our $r4 = qr{
   \(
     (
       [^()]*
       |
       (??{ our $r4 })
     )*
   \)
   }xms;
 ;;
 @p = $s =~ m{ $r4 }xmsg;
 print qq{'$_'} for @p;
 print '--------';
 ;;
 $s =~ m{ ($r4) [^()]* ($r4) }xms;
 print qq{1 '$1'   3 '$3'};
 print '--------';
"
'(y)'
'(a(b))'
'()'
'()'
'(a(b)c()(d(e(f)g))h)'
--------
1 '(y)'   2 '(a(b))'
--------
''
''
''
''
''
--------
1 '(y)'   3 '(a(b))'
--------
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Have you tried the very neat  "(?PARNO)" operator available with Perl 5.10+ and discussed in the example referred to here?

"... our = our"

Yes, I am familiar with the reflexive property of equality. (For all X, X == X.) Can you be a little less enigmatic? Are you saying I cannot use the "our" scope? When I used "my" I created an entirely separate $paren and broke the recursion, and because I have "use strict" I cannot leave the variable undeclared. Perhaps I just cannot use "use strict"?

... our is our

I think Anonymonk is concerned about the second appearance of  our $re within the  (??{ our $re }) regex subexpression. Although somewhat unfamiliar, I think this is kosher because our acts as a declaration of a package variable and  $re needs to be declared or pre-declared somehow to be referenced within the regex under full strictures. The following variations work identically with strictures (note no capturing groups within $re):

>perl -wMstrict -le
"my $s = 'x(y) (a(b)) ()() q (a(b)c()(d(e(f)g))h) q';
 ;;
 our $r3;
 $r3 = qr{
   \(
     (?:
       [^()]+
       |
       (??{ $r3 })
     )*
   \)
   }xms;
 ;;
 my @p = $s =~ m{ $r3 }xmsg;
 print qq{'$_'} for @p;
 print '--------';
 ;;
 $s =~ m{ ($r3) [^()]* ($r3) }xms;
 print qq{1 '$1'   2 '$2'};
 print '--------';
"
'(y)'
'(a(b))'
'()'
'()'
'(a(b)c()(d(e(f)g))h)'
--------
1 '(y)'   2 '(a(b))'
--------
[download]

And the OPed:

our $r3 = qr{
  ...       
    (??{ our $r3 })
  ...
  }xms;
[download]