## Re: [PrimeNumbers] Interesting Sets of Three Consecutive Primes

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• ... 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
Message 1 of 4 , Sep 10, 2005
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
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