Useful facts:

1. If a prime p divides n, then
	phi(p n) = p phi(n);
moreover,
	phi(p^k n) = p^k phi(n).
Further, if every prime divisor of m also divides n, then
	phi(m n) = m phi(n).

2. For n > 1,
	sigma(p^n) - 1 = p sigma(p^(n-1));
further, for k <= n,
	sigma(p^n) - sigma(p^(k-1)) = p^k sigma(p^(n-k)).


Simple transformation theorem:

Suppose N = phi(sigma(N)), p^k || N, and let N' = N/p^k.  Then
p N = phi(sigma(p N)) if the following conditions hold:

	a. sigma(p^(k+1)) = P, a prime not dividing sigma(N');
	b. every prime dividing sigma(p^k) also divides sigma(N').

Proof.

	N = phi(sigma(N))
	  = phi(sigma(N' p^k))
	  = phi(sigma(N') sigma(p^k))
	  = phi(sigma(N')) sigma(p^k).

	phi(sigma(p N)) = phi(sigma(N' p^(k+1)))
				 = phi(sigma(N') sigma(p^(k+1)))
				 = phi(sigma(N') P)
				 = phi(sigma(N')) (P - 1)
				 = phi(sigma(N')) (sigma(p^(k+1)) - 1)
				 = phi(sigma(N')) p sigma(p^k)
				 = p N.	QED


Generalized transformation theorem:

Suppose N = phi(sigma(N)), p^k || N, and let N' = N/p^k.  Then
p^r N = phi(sigma(p^r N)) if the following conditions hold:

	a. sigma(p^(k+r))/sigma(p^(r-1)) = P, a prime not dividing sigma(N');
	b. every prime dividing sigma(p^k)/sigma(p^(r-1)) also divides sigma(N').

Notes:  This reduces to the simple transformation theorem when r = 1.
	The quotients are integers iff r | (k + 1); P being prime further
	requires (k + r + 1)/r to be prime.

Proof.  We have
	N = phi(sigma(N))
	  = phi(sigma(N' p^k))
	  = phi(sigma(N') sigma(p^k))
	  = phi(sigma(N') sigma(p^(r-1))) sigma(p^k)/sigma(p^(r-1)).

	phi(sigma(p^r N)) = phi(sigma(N') sigma(p^(k+r)))
			= phi(sigma(N') sigma(p^(r-1)) P)
			= phi(sigma(N') sigma(p^(r-1))) (P - 1)
			= phi(sigma(N') sigma(p^(r-1))) (sigma(p^(k+r))/sigma(p^(r-1)) - 1)
			= phi(sigma(N') sigma(p^(r-1))) p^r sigma(p^k)/sigma(p^(r-1))
			= p^r N. 	QED

Remark:  We know P does not divide sigma(p^(r-1)) because P is larger.


Special transformation theorem:

Suppose N = phi(sigma(N)), p^k || N, and let N' = N/p^k.  If the following
conditions hold:

	a. sigma(p^(k+1)) = P, a prime not dividing sigma(N');
	b. every prime but one (say q) dividing sigma(p^k) also divides sigma(N');
	c. q divides sigma(p^k) only once and does not divide N;
	d. (q + 1)/2 = q', a prime not dividing sigma(N');
	e. sigma(N') is even;

then p q N = phi(sigma(p q N)).

Proof.  We have
	N = phi(sigma(N))
	  = phi(sigma(N' p^k))
	  = phi(sigma(N') sigma(p^k))
	  = phi(sigma(N') q) sigma(p^k)/q
	  = phi(sigma(N')) (q - 1) sigma(p^k)/q.

	phi(sigma(p q N)) = phi(sigma(N' p^(k+1) q))
				 = phi(sigma(N') sigma(p^(k+1)) sigma(q))
				 = phi(sigma(N') P 2 q')
				 = phi(sigma(N')) (P - 1) 2 (q' - 1)
				 = phi(sigma(N')) p sigma(p^k) (q - 1)
				 = p q N. 	QED