Theorem on the k differences between two primes
                                                           By Cino Hilliard
                                                               Jan 9, 2005 

Theorem 1

If p+k is prime then p+k divides p^(p+k) + k. 

We use Mathematical Induction and the Binomial Theorem to prove this. 

First we use induction to prove

Lemma 1:  p^(p+k) - p is divisible by p+k if p+k is prime. 

For p=1 we have 1^(1+k) + k = 1+k and 1+k divides 1+k. So the statement is true for 1.

Induction hypothesis: Assume p^(p+k) - p  is divisible by p+k for prime p+k.

Since we assume p+k is prime, the Binomial Theorem expansion of ( p+1)^(p+k) gives us 

(p+1)^(p+k) = p^(p+k) + (p+k)hp + 1 for some integer h. Then

(p+1)^(p+k) - p^(p+k) - 1 = (p+k)hp.  Now by adding p^(p+k) - p to both sides we get

 1.  (p+1)^(p+k) - (p+1) = (p+k)hp + p^(p+k) - p and using the induction hypothesis,

 2.  (p+1)^(p+k) - (p+1) = (p+k)hp + (p+k)m   for some m. 

Since p+k divides the right side of 2.  it divides the left. Thus we have shown that p^(p+k) - p is

divisible by p+k for p+1 and by the induction  p^(p+k) - p is divisible by p+k all p if p+k is prime. 

Proof  I  
3.  p^(p+k) + k = p^(p+k) - p + p + k  = p^(p+k)-p +(p+k)     (The trick from Broadhurst -p+p)

Then by Lemma 1.,  p^(p+k) - p = (p+k)m for some m. Substituting this into 3, we have

p^(p+k) + k = (p+k)m + (p+k) which implies p^(p+k) + k is divisible by p+k.
QED

Proof II
Lemma 2.
For p < q and q prime, p^q + (q-p) is divisible by q.  
p^q - p is div by q by Lemma 1 so p^q - p + q is div by q as desired.

So for Theorem 1  If p+k is prime then p+k divides p^(p+k) + k.

we let p+k = q in Lemma 2. Thus Lemma 2 becomes

p^(p+k) - p + p+k is divisible by p+k or

p^(p+k) + k is divisible by p+k if p+k is prime as desired.

Further, if  p+k is not prime then at least one factor of p+k divides p^(p+k) + k.