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