- I am looking for all twin primes such that their arithmetic mean is a

perfect number.

For example:

If 5 and 7 are our primes then 6 is our arithmetic mean.

Thank you for your help. - --- In primenumbers@y..., "math4life2000" <Math4life@a...> wrote:
> I am looking for all twin primes such that their arithmetic mean is

a

> perfect number.

I assume that you mean all twin prime pairs (p, p+2) such that

>

> For example:

>

> If 5 and 7 are our primes then 6 is our arithmetic mean.

>

> Thank you for your help.

p+1 is a perfect number. If that is your intent, you found the

only such pair already. All perfect numbers greater than 6 are

congruent to 1 (modulo 3), meaning that the integer immediately

before the perfect number is divisible by 3 (making it non-prime).

On the other hand, if the two primes don't have to be part of

a single twin prime pair, there could be many solutions. The next

solution after (5,7) would be (13,43). - --- In primenumbers@y..., "jbrennen" <jack@b...> wrote:

> All perfect numbers greater than 6 are congruent to 1 (modulo 3)

Before somebody corrects me on this, let me rephrase:

All EVEN perfect numbers greater than 6 are congruent

to 1 (modulo 3).

Obviously, an odd perfect number (if such exists) cannot be the

number in the middle of a twin prime pair.

As a side note, there are probably no odd perfect numbers, so

the original statement was probably true, but until somebody

proves the nonexistence of odd perfect numbers, I must watch

myself :-) - On Mon, 01 Apr 2002 21:38:27 -0000, you wrote:

>I am looking for all twin primes such that their arithmetic mean is a

There aren't any others.

>perfect number.

>

>For example:

>

>If 5 and 7 are our primes then 6 is our arithmetic mean.

>

>Thank you for your help.

[Apologies to the real mathematicians for the standard of the proof. I

hope the logic is clear]

Given that these 3 numbers must be adjacent integers, then one of them

must be divisible by 3. Neither of the primes is 3 (because no number

adjacent to this is perfect), so it must be the perfect number that is

divisible by 3. It must be an even perfect number (because otherwise the

primes would both be even) and is therefore the product of a power of

two and a Mersenne prime. Obviously a power of two is not divisible by

three and so the Mersenne prime must be 3. And the only perfect number

of this form is 6, which gives you the solution you already have.

Regards

Steve nice but dim