Bill Sindelar wrote:

> Jens thank you for your quick response. But you have me stumped. Can I

> interpret your response as follows?

> Yes, there exist sets of 3 CONSECUTIVE primes P, Q and R, such that the

> PRODUCT (2*Q* R) is equal to the SUM of 2 distinct PRIMES S<T, whose

> DIFFERENCE (T-S) is divisible by the SUM (Q+R) and yields a QUOTIENT

> WHICH EQUALS P. However, if the 3 consecutive primes are in arithmetical

> progression, with the sole exception of P=3, Q=5, R=7, then no such sets

> can exist.

Yes, that is correctly interpreted. My formulation:

"There are cases of consecutive primes P, Q, R such that Q*R +/- P*(Q+R)/2 are

both prime"

is simply a rewording of your slightly convoluted:

"There exist ..... which equals P".

You can check this by setting S, T = Q*R +/- P*(Q+R)/2.

P, Q, R = 10^100+8556049 + 0, 78, 464 is a solution found in a few minutes

with PARI/GP.

Heuristics support infinitely many solutions (reminds of twin prime

conjecture) but I have no idea how to prove it.

If P, Q, R, is an AP3 solution other than 3, 5, 7, then your S cannot be prime

since 3 always divides it when S and T meet the other conditions.

--

Jens Kruse Andersen