* A square such that area of inner region A equals area of region B:
   __________________
  |             :    |
  |             :    |
  |             :    |
  |      A      :    |
  |             :    |
  |             :    |
  |.............:    |
  |               B  |
  |__________________|

  Also, what about "rings" such that the area of A equals that of B
  (see below)?  This would extend better to other shapes, such as
  circles.  Yeah, try circles first.  Wait a minute, the answer ought
  to be obvious... oh, duh, it's just when one circumference is twice
  as long as the other, since they're just multiplied by a constant
  pi.  But the same is not true of the other shapes, aha!

   __________________
  |  ,............,  |
  |  :            :  |
  |  :            :  |
  |  :     A      :  |
  |  :            :  |
  |  :            :  |
  |B :............:  |
  |__________________|


* (N + 1)^2 = N^2 + 2N + 1

        and

  (N + 1)^3 = N^3 + (N + 1)^2 + 2N^2 + N

  Visualize a square, then a cube, to see why.  Now go on to
  hypercube, then n-cubes. 


* The number of N's divisors is equal to the number of unique
  multiplicative combinations of its unique prime factors?  Let's see:

  Instead of generating primes, how about generating composites and
  taking the holes?

* Some thoughts:
  Def:    Sum-Pfactrs(N) = the sum of its prime factorization
  Def:    Choiceness(N) = N / Sum-Pfactrs(N)

  Or maybe choiceness is more appropriately the number of prime
  factors (with 1 being highest grade of choice, also called prime),
  or maybe unique prime factors, or...

  How does Sum-Pfactrs(N) compare with the sum N's divisors... compare
  with the sum of [1...N]?


* Are there any odd perfect numbers?  A perfect number is the
  sum of its divisors, including 1 but not including itself.  For
  example: 6 = 1 + 2 + 3.

  Thoughts: well, let's take a look at divisors of odd and even
  numbers, and see if we can't find some pattern.  Then see if the
  pattern admits of proof?

  Each divisor D1 of a number N has a "partner" D2, such that (D1 * D2
  = N), right?  Here are two parity tables, where E means even and O
  means odd:

   (D1 * D2 = N)                       (D1 + D2 = N)
    _________________                   _________________ 
   |N     D1     D2  |                 |N     D1     D2  |
   |-----------------|                 |-----------------|
   |E     E     O/E  |                 |E     E      E   |
   |      O      E   |                 |      O      O   |
   |.................|                 |.................|
   |O     E      ?   |                 |O     E      O   |
   |      O      O   |                 |      O      E   |
   |_________________|                 |_________________|


  The only ambiguity is in the first table; if N and D1 are both even,
  then D2 can be even or odd, and if N is odd, then neither D1 nor D2
  can be even.

  Forgetting about the 1, since it's a constant component anyway, we
  know that any number N has an even number of divisors, because each
  divisor D1 must have its counterpart D2.  No D1 could have two or
  more different counterparts, for (N / D1) must always equal the same
  number.

  So what would it mean for an odd number N to be perfect?  Well, it
  would mean that an even number of divisors had summed up to an even
  number (remember, it will be odd once we add the 1 back on).  If X
  of those divisors are odd, then X must be even, for N to be even as
  well.  By definition, all the other divisors of N are even numbers,
  if they exist at all.

  This means that an even number of even numbers has been added to an
  even number of odd numbers, to produce an even number.  Then 1 is
  added, to produce an odd number.  All well and good, no
  contradictions so far.

  Now let's multiply them out instead of adding.  We should obtain the
  same N, but we can really forget about the 1 this time.  If N is to
  be odd, then it can have no even divisors at all (it only takes one
  of them to "contaminate" the product).  It follows that all the
  divisors of N must be odd, and there must be an even number of them.
  This leads to the first requirement for an odd perfect number N:

  Req. 1: All divisors of N are odd.

  This could have been arrived at much more simply, by noting that any
  even divisors at all would lead to an even N.  Sigh.

  IDEA: wait!!!  There is one thing about even Ns -- they all, without
  fail, have 2 as a divisor.  There is no such required number for odd
  Ns -- any old collection of odd numbers willl do.  Hmm... can a
  contradiction be shown in having your divisors sum up to yourself if
  they are all greater than 2 (except for the 1)?  Seems the most 
  promising way to go.


* Hey, there's always Goldbach's conjecture. :-)