- This exercise stems from my previous post "A Remarkable Property of

Consecutive Integer Pairs; Odd, Even and Prime." Anyone care to carry it

further?

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. Here is the smallest and largest examples I was able to

find out of a total of 12:

P=3, Q=5, R=7, (2*Q*R)=70, S=17, T=53, (T-S)=36, (Q+R)=12, quotient=3

which equals P.

P=1039, Q=1049, R=1051, (2*Q*R)=2204998, S=11549, T=2193449,

(T-S)=2181900, (Q+R)=2100, quotient=1039 which equals P.

I looked for sets of 3 consecutive primes in arithmetic progression but

found only one so far, which is the first example shown above. Maybe more

3-PAP's of this type cannot exist and that sets of 3 consecutive primes

of this type are finite in number. I'm without a clue. I would appreciate

comments. Thanks folks.

Bill Sindelar - Bill Sindelar wrote:

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

In other words:

> 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.

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

both prime.

If P, Q, R is a prime AP3 with P>3 then P, Q, R are either all 1 or all 2

modulo 3.

In both cases 3 divides Q*R - P*(Q+R)/2.

--

Jens Kruse Andersen - On Thu, 8 Sep 2005 23:28:10 +0200 "Jens Kruse Andersen"

<jens.k.a@...> writes:> Bill Sindelar wrote:

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

>

> > 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.

>

> In other words:

> There are cases of consecutive primes P, Q, R such that Q*R +/-

> P*(Q+R)/2 are

> both prime.

>

> If P, Q, R is a prime AP3 with P>3 then P, Q, R are either all 1 or

> all 2

> modulo 3.

> In both cases 3 divides Q*R - P*(Q+R)/2.

>

> --

> Jens Kruse Andersen

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.

Thanks again and regards.

Bill Sindelar - Bill Sindelar wrote:

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

Yes, 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

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