## New conjecture

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• Let a and b be two relatively prime numbers. f(a,b) = a^2 - b^2 - a*b The conjecture is that: f(a,b) = 1mod10 or (-1)mod 10 or 0mod5. and that: f(a,b) will
Message 1 of 8 , Jun 30, 2001
Let a and b be two relatively prime numbers. f(a,b) = a^2 - b^2 - a*b The
conjecture is that: f(a,b) = 1mod10 or (-1)mod 10 or 0mod5. and that:
f(a,b) will generate the set of every prime number of the form (+/-1)mod10
and the prime number 5 itself. There are an infinity of a and b which will
generate each member of this set and that: f(a,b) also generates are all
the possible composites of the prime numbers contained in the above set.

I have a web site skywebsite.com/mistermac/mistermac/ which contains the
above conjecture.
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• ... [....] First part of conjecture is easily proven, using the fact that 10 is an even number. For a=1 to 10, for b=1 to 10, compute w=f(a,b)%10 and discard
Message 2 of 8 , Jun 30, 2001
John McNamara wrote:

> Let a and b be two relatively prime numbers.
> f(a,b) = a^2 - b^2 - a*b
> The conjecture is that:
> f(a,b) = 1 mod 10 or (-1) mod 10 or 0 mod 5.
> and that:
[....]

First part of conjecture is easily proven,
using the fact that 10 is an even number.

For a=1 to 10, for b=1 to 10, compute w=f(a,b)%10
By inspection, the remaining cases have
a and b both even, and hence not coprime.
So there are no exceptions in these 100 cases.
But then by adding arbitrary multiples of 10
we prove the conjecture for all coprime a and b,
since 10 is an even number.

Second part

> f(a,b) will generate the set of every prime number of the form
> (+/-1)mod10
[...]

This is way too hard for me to comment usefully on, sorry.

David
• Well here s an obviously necessary condition. Let p be any prime such that p^2=1 mod 10. Then the conjecture requires the existence of an integer b such that
Message 3 of 8 , Jun 30, 2001
Well here's an obviously necessary condition.
Let p be any prime such that p^2=1 mod 10.
Then the conjecture requires the
existence of an integer b such that 5*b^2+4*p
is a perfect square [so that (2*a-b)^2=5*b^2+4*p]
I checked that for all p < 5,000,000.
But it needs someone skilled in such (Pellian?)
diophantine work to show that this is always the case.
There is little point in looking for a counterexample,
since the condition is very easy to satisfy, in practice.
[E.g: b=10^3-1, for p=5*10^6-1]
I have the strong feeling that a good number theorist
[not me!] could prove such a thing. I seem to recall that
Pellian equations are solved by Lucas sequences,
so maybe Marcel can prove it?
• I haven t seen this conjecture before: There are an infinite number of primes in sigma(n=1,k,p^k). i.e. 1 1+p 1+p+p^2 1+p+p^2+p^3 1+p+p^2+p^3+p^4 etc...
Message 4 of 8 , May 22, 2002
I haven't seen this conjecture before:

There are an infinite number of primes in sigma(n=1,k,p^k).

i.e.

1
1+p
1+p+p^2
1+p+p^2+p^3
1+p+p^2+p^3+p^4
etc...

contains an infinite number of prime terms.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• ... Jon, You may not have seen it, but I m sure that you are familiar with the names Caldwell and Dubner? See:
Message 5 of 8 , May 22, 2002
--- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
> I haven't seen this conjecture before:
>
> There are an infinite number of primes in sigma(n=1,k,p^k).
>
> i.e.
>
> 1
> 1+p
> 1+p+p^2
> 1+p+p^2+p^3
> 1+p+p^2+p^3+p^4
> etc...
>
> contains an infinite number of prime terms.

Jon,

You may not have seen it, but I'm sure that you are familiar
with the names Caldwell and Dubner? See:

http://www.utm.edu/~caldwell/preprints/unique.pdf

On page 4 of that preprint:

"Are there infinitely many repunit primes base b
for every base b?"

... "it seems the [answer] to [this question] should be ... yes"
• ... Probably quite difficult to prove of course; proving this would include the lesser task of proving the infinitude of Mersenne primes, and we know that
Message 6 of 8 , May 22, 2002
--- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
>
> [A conjecture...]
> 1
> 1+p
> 1+p+p^2
> 1+p+p^2+p^3
> 1+p+p^2+p^3+p^4
> etc...
>
> contains an infinite number of prime terms.

Probably quite difficult to prove of course; proving this would
include the lesser task of proving the infinitude of Mersenne
primes, and we know that isn't easy to prove.

It is easy to show that the repunits in a given base (i.e., the
sequence enumerated above) cannot have a finite covering set
of primes.
• Page 3 of that PDFprint: probable primes are postive integers p which satisfy Fermat s congruence a^p=1modp. And the rest..... Jon Perry perry@globalnet.co.uk
Message 7 of 8 , May 22, 2002
Page 3 of that PDFprint:

probable primes are postive integers p which satisfy Fermat's congruence
a^p=1modp.

And the rest.....

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: jbrennen [mailto:jack@...]
Sent: 22 May 2002 19:35

--- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
> I haven't seen this conjecture before:
>
> There are an infinite number of primes in sigma(n=1,k,p^k).
>
> i.e.
>
> 1
> 1+p
> 1+p+p^2
> 1+p+p^2+p^3
> 1+p+p^2+p^3+p^4
> etc...
>
> contains an infinite number of prime terms.

Jon,

You may not have seen it, but I'm sure that you are familiar
with the names Caldwell and Dubner? See:

http://www.utm.edu/~caldwell/preprints/unique.pdf

On page 4 of that preprint:

"Are there infinitely many repunit primes base b
for every base b?"

... "it seems the [answer] to [this question] should be ... yes"

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• ... Very true - in fact see: http://www.users.globalnet.co.uk/~perry/maths/uniqueantidivisor.htm#infinite primes I was hoping this would allow us to see the
Message 8 of 8 , May 22, 2002
>It is easy to show that the repunits in a given base (i.e., the
>sequence enumerated above) cannot have a finite covering set
>of primes.

Very true - in fact see:

http://www.users.globalnet.co.uk/~perry/maths/uniqueantidivisor.htm#infinite
primes

I was hoping this would allow us to see the true differential between
Mersenne's (in which the sums do sum to 2^k-1), and other similar sequences
(which don't (sum to 2^k-1 or p^k-1)).

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: jbrennen [mailto:jack@...]
Sent: 22 May 2002 19:55

--- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
>
> [A conjecture...]
> 1
> 1+p
> 1+p+p^2
> 1+p+p^2+p^3
> 1+p+p^2+p^3+p^4
> etc...
>
> contains an infinite number of prime terms.

Probably quite difficult to prove of course; proving this would
include the lesser task of proving the infinitude of Mersenne
primes, and we know that isn't easy to prove.

It is easy to show that the repunits in a given base (i.e., the
sequence enumerated above) cannot have a finite covering set
of primes.

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