# Using your Head is Permitted

## May 2009 solution

The sum of the *proper* divisors of a perfect number is itself, so
the sum of *all* of its divisors is twice itself. We will denote
the sum of all divisors of a number, *n*, as σ(*n*).
Suppose that a number *n* is perfect. It can be represented by its
factorization as a product of *p*_{i}^*k*_{i}, where
the *p*_{i} are a finite set of distinct primes. The sum of the
divisors of *p*_{i}^*k*_{i} is
*p*_{i}^*k*_{i}+*p*_{i}^(*k*_{i}-1)+ ... +1,
and it can easily be shown that σ(*n*) is the product of these sums
for all of *n*'s prime factors (because if *a* and *b* are
co-prime, σ(*ab*)=σ(*a*)σ(*b*) ).

For *n* to be perfect, σ(*n*)=2*n*. Equivalently, the
product of (*p*_{i}^*k*_{i}+*p*_{i}^(*k*_{i}-1)+ ... +1)/*p*_{i}^*k*_{i}
for all *i* must equal 2.

If *n* is odd, so are all denominators in this product. This means
that all numerators must also be odd, except for one that is 2 (mod 4). Because
all *p*_{i} are odd, all powers of *p*_{i}
are also odd, and the parity of *p*_{i}^*k*_{i}+*p*_{i}^(*k*_{i}-1)+ ... +1
is simply the parity of the number of summed elements, meaning that for it to
be odd, *k*_{i} must be even. These *p*_{i} values
contribute to the "(2*m*+1)^{2}" part of the
representation, because the corresponding
*p*_{i}^*k*_{i} values are odd squares. The remaining
*p*_{i} equals "*q*". Its
*k*_{i} must be odd. Its
*p*_{i}^(*k*_{i}-1) part, being an odd square,
completes the "(2*m*+1)^{2}" part of the representation.

In order to show that this is special form B, we still need to prove that
*q* is congruent to 1 (mod 4). The reason for this is that otherwise the
numerator for this particular *p*_{i} would have been 0 (mod 4),
contrary to the 2 (mod 4) assumption.

In contrast, *n* may be even. If it is, consider it as
2^{s}*t*, with odd *t*. Let *p*_{0}=2,
then the numerator
for this prime is 2^{s+1}-1, which we shall denote *u*.
All prime factors of *u* must appear in the *p*_{i} list (for
their values in the denominator to cancel out the odd value that appears in the
numerator), so the smallest odd *p*_{i} must be smaller or equal
to *u*. Let us call it *p*. Note that if *p*=*u* then
2^{s}*u* is a perfect number. This indicates
*t*=*u*, because adding any further prime is adding another multiple
that is larger than one to the product that calculates
σ(*n*)/*n*, so the total product will be larger than 2. We can
also not increase the power of *p* from 1, because
*p*^{k}+*p*^{k-1}+...+1>*p*+1 for
*k*>1, once again pushing the product above 2.

On the other hand, *p* can not be smaller than *u*, for this would
indicate (*p*+1)/*p*>(*u*+1)/*u*, again pushing the
product above 2. As before, adding more primes or increasing the power of
*p* merely makes this problem worse.

Therefore, *t*=*u* must be a prime, so we conclude that
*n* must be of special form A.

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