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• Hi all Have been looking at a^r + b^s = prime, where a,b are adjacent integers as are r,s. Occassionally there is a glimmer of order, like for instance 20^3 +
Message 1 of 6 , Dec 17, 2005
Hi all

Have been looking at a^r + b^s = prime,
where a,b are adjacent integers as are r,s.

Occassionally there is a glimmer of order, like for instance

20^3 + 21^4 = prime
21^3 + 22^4 = prime
22^3 + 23^4 = prime
23^3 + 24^4 = prime
24^3 + 25^4 = prime.

Is there any longer? (I don't know, haven't looked into it.)

Anyways I wonder, given any integer a >1 is there always at least one
adjacent integer b such that a^r + b^s is prime, where r,s are
adjacent integers no more than a?

For instance:

26: 26^2 + 27^3 = prime
29: 29^10 + 28^9 = prime
35: 35^2 + 34^3 = prime

Mark
• Hi Mark, odd prime numbers have the final number 1,3,7 or 9. The expression n^3 + (n+1)^4 is, as easily to show, always odd and not divisible by 5, and thus
Message 2 of 6 , Dec 18, 2005
Hi Mark,

odd prime numbers have the final number 1,3,7 or 9. The expression
n^3 + (n+1)^4 is, as easily to show, always odd and not divisible by
5, and thus has likewise the final numbers 1,3,7 or 9. n and (n+1)
are coprime, their powers too, and the sum of the powers is thus not
divisible by the prime factors of n and (n+1). All this creates a
certain predisposition for primality. But only a certain.

Werner
• From: Werner D. Sand Date: 12/18/05 13:52:38 To: primenumbers@yahoogroups.com Subject: [PrimeNumbers] re:adjacents and primes Hi Mark, odd prime numbers have
Message 3 of 6 , Dec 18, 2005
From: Werner D. Sand
Date: 12/18/05 13:52:38

Hi Mark,

odd prime numbers have the final number 1,3,7 or 9. The expression
n^3 + (n+1)^4 is, as easily to show, always odd and not divisible by
5, and thus has likewise the final numbers 1,3,7 or 9. n and (n+1)
are coprime, their powers too, and the sum of the powers is thus not
divisible by the prime factors of n and (n+1). All this creates a
certain predisposition for primality. But only a certain.

Werner

From kermit@...

In fact n^3 + (n+1)^4

is never divisible by 3, by 5, or by 31, and maybe not divisible by some
few primes greater than 31.

Note that 31 is equal to 1 mod 3 and equal to 1 mod 5.

n^3 + (n+1)^4 can be divisible by 7, 11, 13, 17, 23 and 29.

[Non-text portions of this message have been removed]
• From: Mark Underwood Date: 12/17/05 13:21:44 To: primenumbers@yahoogroups.com Subject: [PrimeNumbers] adjacents and primes Hi all Have been looking at a^r +
Message 4 of 6 , Dec 18, 2005
From: Mark Underwood
Date: 12/17/05 13:21:44

Hi all

Have been looking at a^r + b^s = prime,
where a,b are adjacent integers as are r,s.

From kermmit@...

This is equivalent to investigating

whether or not

a^r + (a+1)^(r+1) is a prime polynomial.

Generally prime polynomials produce prime integers in the pattern
you have observed.

[Non-text portions of this message have been removed]
• ... 7 divides a^3 + (a+1)^4 for a==5 (mod 7). 6 consecutive primes is then the best possible for that formula. It first occurs starting at a = 841369395. a^5 +
Message 5 of 6 , Jan 3, 2006
December 17 Mark Underwood wrote:

> Have been looking at a^r + b^s = prime,
> where a,b are adjacent integers as are r,s.
>
> Occassionally there is a glimmer of order, like for instance
>
> 20^3 + 21^4 = prime
> 21^3 + 22^4 = prime
> 22^3 + 23^4 = prime
> 23^3 + 24^4 = prime
> 24^3 + 25^4 = prime.
>
> Is there any longer? (I don't know, haven't looked into it.)

7 divides a^3 + (a+1)^4 for a==5 (mod 7).
6 consecutive primes is then the best possible for that formula.
It first occurs starting at a = 841369395.

a^5 + (a+1)^4 never has a factor below 17.
17 divides for both a==1 and a==4 (mod 17).
The formula can avoid small factors for 13 consecutive a.
The first case of 6 primes starts at a = 1469217.
The first 7 is:

188934537^5 + 188934538^4
188934538^5 + 188934539^4
188934539^5 + 188934540^4
188934540^5 + 188934541^4
188934541^5 + 188934542^4
188934542^5 + 188934543^4
188934543^5 + 188934544^4

> Anyways I wonder, given any integer a >1 is there always at least one
> adjacent integer b such that a^r + b^s is prime, where r,s are
> adjacent integers no more than a?
>
> For instance:
>
> 26: 26^2 + 27^3 = prime
> 29: 29^10 + 28^9 = prime
> 35: 35^2 + 34^3 = prime

I don't allow exponent 0 (a^0=1).

If exponent 1 is allowed:
The smallest a>1 for which a^r + b^s is never prime for
b = a+/-1, s = r+/-1, and r,s <= a, is a = 13361.
This assumes prp's for smaller a are really primes.

The cases where prp's above 3000 digits were needed:
1252^1155 + 1253^1154 (3578 digits)
3319^992 + 3318^991 (3493 digits)
9818^765 + 9819^764 (3054 digits)
9819^764 + 9818^765 (3054 digits, same prp as for a=9818)
10127^888 + 10126^887 (3557 digits)
11051^1176 + 11050^1177 (4760 digits)

PrimeForm/GW made prp tests. 13361^13361 has 55126 digits.
I stopped there but guess there are primes for larger exponents.

If exponent 1 is not allowed then the prime 3743^1 + 3744^2 is
skipped and the smallest a>2 is a = 3743.
For a=2, there are no allowed r,s values at all.

--
Jens Kruse Andersen
• Jens I don t know how you can go up that high, good lord! Amazing. Thank you for finding (at over 55,000 digits!) what I could never have hoped to. kind
Message 6 of 6 , Jan 3, 2006
Jens I don't know how you can go up that high, good lord! Amazing.
Thank you for finding (at over 55,000 digits!) what I could never
have hoped to.

kind regards,
Mark

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@g...> wrote:
>
(snip)

> The smallest a>1 for which a^r + b^s is never prime for
> b = a+/-1, s = r+/-1, and r,s <= a, is a = 13361.
> This assumes prp's for smaller a are really primes.
>
> The cases where prp's above 3000 digits were needed:
> 1252^1155 + 1253^1154 (3578 digits)
> 3319^992 + 3318^991 (3493 digits)
> 9818^765 + 9819^764 (3054 digits)
> 9819^764 + 9818^765 (3054 digits, same prp as for a=9818)
> 10127^888 + 10126^887 (3557 digits)
> 11051^1176 + 11050^1177 (4760 digits)
>
> PrimeForm/GW made prp tests. 13361^13361 has 55126 digits.
> I stopped there but guess there are primes for larger exponents.
>
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