Conference rusure::math

>"Prove that the sum of the divisors of an integer n (including 1 and n itself)
>is odd if and only n is a square or twice a square."

For the moment assume n is odd, thus has only odd divisors. For each 
divisor there is a corresponding quotient. If n is not a square then all the 
divisors and quotients are paired; the pairs of divisors and quotients -
which are also divisors - sum to an even number. 

However, if n is square = N^2 then N has no opposite to pair off with. All 
the other pairs of divisors contribute an even parity to the sum, but N is
odd, so the sum is odd. 

If we add powers of 2 to the divisors of n, it has no effect on the parity
of the sum of n's divisors, so we are left with the result as stated.

Lynn Yarbrough 

    Let N = 2^A * (p ^ e ) * (p ^ e ) * ... * (p ^ e ),
                    1   1      2   2            m   m
    
    where A >= 0, e  > 0, and the p  are distinct odd primes.
                   i               i
    
    Then N has (A+1) * (e + 1) * (e + 1) * ... * (e + 1) distinct divisors,
                         1         2               m
    
        2^f  * (p ^ f ) * (p ^ f ) * ... * (p ^ f ),
           0     1   1      2   2            m   m
    
    where 0 <= f  <= A, and 0 <= f  <= e .
                0                 i     i
    
    Of these, there are (e + 1) * (e + 1) * ... * (e + 1) odd divisors of N.
                          1         2               m
    
    Let S(N) be the sum of the divisors of N (including 1 and N).  S(N) is odd
    iff N has an odd number of odd divisors.  But the number of odd divisors
    is (e + 1) * (e + 1) * ... * (e + 1), which is odd iff each e  is even.
         1         2               m                             i
    
    Thus, S(N) is odd iff N = 2^a * Q, where Q is a product of even powers of
    odd primes; i.e., Q is an odd square.
    
    [I'm unhappy with the rest of this proof -- it's not very clean]
    
    Let Q = q^2, q is odd.
    
    If A is even, then S(N) is odd iff N = 2^A * Q = (2^a * q)^2, where 2*a = A.
    If A is odd, S(N) is odd iff N = 2^A * Q = 2 * (2^b * q)^2, where 2*b+1 = A.
    
    Since A must either be odd or even, this covers all possibilities, and so
    S(N) is odd iff N is a square or twice a square.

     What's wrong with "the rest of this proof"?
     
     You have shown that the "odd part" of the number is a
     square.  This leaves you with either a square or twice a
     square.  The next power of two gives you four times a
     square, which is a square.  Eight times a square is twice a
     square.  Etc.
     
     Dan

Nothing's *wrong* with it -- it's awkward.  I had trouble expressing the rest
of the argument clearly, and swept much of the tedious stuff 'under the rug'.

For example, showing that:

      If (N = r^2, for some integer r) or (N = 2*r^2, for some integer r)
      then N can be written in the form N = 2^a * Q, where Q is an odd square.

      If NOT ((N = r^2) or (N = 2*r^2)) then N canNOT be written in the form....

Yes, it's obvious, but I don't see a succinct way to show that the two 'forms'
are equivalent.