Re: [PrimeNumbers] Interesting Sets of Three Consecutive Primes
- Bill Sindelar wrote:
> Jens thank you for your quick response. But you have me stumped. Can IYes, that is correctly interpreted. My formulation:
> 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.
"There are cases of consecutive primes P, Q, R such that Q*R +/- P*(Q+R)/2 are
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
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