## Interesting Sets of Three Consecutive Primes

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• 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
Message 1 of 4 , Sep 8, 2005
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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
Bill Sindelar
• ... 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
Message 2 of 4 , Sep 8, 2005
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Bill Sindelar wrote:

> 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
• On Thu, 8 Sep 2005 23:28:10 +0200 Jens Kruse Andersen ... Jens thank you for your quick response. But you have me stumped. Can I interpret your response as
Message 3 of 4 , Sep 10, 2005
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On Thu, 8 Sep 2005 23:28:10 +0200 "Jens Kruse Andersen"
<jens.k.a@...> writes:
> Bill Sindelar wrote:
>
> > 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

Jens thank you for your quick response. But you have me stumped. Can I
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
• ... 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 4 of 4 , Sep 10, 2005
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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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