Prove z-y divides z+y if and only if z=y+2

Theorem I: If z > y are odd primes then z-y divides z+y if and only if z=y+2.

Proof:

Let
(1) z+y = (z-y)k for some integer k.
(2) z=y+2m for some integer m (we must add multiples of 2 to y in order to avoid z=even).

Substituting (2) into (1) we get

y+2m + y = (y+2m-y)k = 2y+2m = 2mk. Then y+m = mk so  y/m + 1 = k for k to be an integer, m  = 1 or m = y. If m = y then k=2 and z+y = 2*(z-y)  = 2z-2y => 3y = z contradicting z is prime.  Therefore, m=1 and z = y+2m = y+2 as desired.

 If z and y are odd primes and  z-y divides z+y  then  y and z are twin primes where y is the lower bound and z is the upper bound.

Proof:
This follows directly from Theorem I for odd primes y and z,  z = y+2 =>  y and y+2 are a twin prime pair.  

Now show if z = y+2 then z-y divides z+y. If z=y+2 then z-y = 2 and z+y = 2y + 2. Since 2 divides 2y+2 the converse is also true.