----- Original Message -----

From: "richard042" <richard042@...>

To: <primenumbers@yahoogroups.com>

Sent: Saturday, August 03, 2002 3:31 PM

Subject: [PrimeNumbers] Re: Proof of D. Andrica conjecture.

> --- In primenumbers@y..., "John W. Nicholson" <johnw.nicholson@a...>

> >

> > Do y'all See any problems?

>

> Maybe,

>

> Bertrand/Tschebycheff: pn<(pn+1)<2*pn

> Take: sqrt(pn) < sqrt(pn+1) < sqrt(2*pn)

> divide by constant c: sqrt(pn)/c < sqrt(pn+1)/c < sqrt(2*pn)/c

>

> A.) subtract lhs: 0 < (sqrt(pn+1)-sqrt(pn))/c < (sqrt(2*pn)-sqrt

> (pn))/c

>

> Regardless of the value of c, A.) must hold.

> We need to get to Andrica's conjecture: sqrt(pn+1) - sqrt(pn)<1.

> So, multiply A.) by c to get:

> B.) 0 < sqrt(pn+1)-sqrt(pn) < sqrt(2*pn)-sqrt(pn)

>

> We can use B.) to prove Andrica's Conjecture only

> if it is always true that rhs <1 - i.e.:

> sqrt(2*pn)-sqrt(pn) < 1

> divide by c=sqrt(pn)to get:

> sqrt(2)-1 < 1/sqrt(pn)

> which is only true for n=1

>

> QESh..

>

> -Dick Boland

Thanks to Dick and David Broadhurst,

You have me doing a change to even a better way of showing this:

Let P(n) = a^2 and P(n+1) = c^2 = (a + b)^2 = a^2 + 2ab + b^2 where a, b,

and c are a real numbers.

Let the gap s^2 = c^2 - a^2 = (c - a)(c + a) = 2ab + b^2 which we assume is

< c + a

Take the square root to find c

c = a + b

s^2 = 2ab + b^2 < a + b + a = 2a + b

Or 2ab + b^2 < 2a + b

So 0 < b < 1

Replace b with y = 1/b so 1 < y however

2a + y must be < a^2 by Bertrand/Tschebycheff:

So y < a^2 - 2a

By adding 1 on the right side we complete the square without going >= a^2

So y <= (a - 1)^2

This is the largest y value; the smallest is y => (a - (a - 1))^2

Let us replace (a-1) with a integer k that make up the complete residue set

for a that is not 0 == mod a.

So there is an integer 1 <= k <= a such that

b = 1/(a - k)^2, c = a + b, and s^2 = 2ab + b^2 < 2a + b

QED

Do I have it this time?- One small but big change See it near bottom.

>

is

> Thanks to Dick and David Broadhurst,

>

> You have me doing a change to even a better way of showing this:

>

> Let P(n) = a^2 and P(n+1) = c^2 = (a + b)^2 = a^2 + 2ab + b^2 where a, b,

> and c are a real numbers.

> Let the gap s^2 = c^2 - a^2 = (c - a)(c + a) = 2ab + b^2 which we assume

> < c + a

Change the following

> Take the square root to find c

> c = a + b

> s^2 = 2ab + b^2 < a + b + a = 2a + b

> Or 2ab + b^2 < 2a + b

> So 0 < b < 1

> Replace b with y = 1/b so 1 < y however

> 2a + y must be < a^2 by Bertrand/Tschebycheff:

> So y < a^2 - 2a

> By adding 1 on the right side we complete the square without going >= a^2

> So y <= (a - 1)^2

> This is the largest y value; the smallest is y => (a - (a - 1))^2

Let us replace (a-1) with a real number k.

So there is an real 0 < k < a such that

b = 1/(a - k)^2, c = a + b, and s^2 = 2ab + b^2 < 2a + b

QED

Do I have it this time? - --- In primenumbers@y..., "John W. Nicholson" <johnw.nicholson@a...>

wrote:

> > Let P(n) = a^2 and P(n+1) = c^2 = (a + b)^2 = a^2 + 2ab + b^2

where a, b,

> > and c are a real numbers.

assume is < c + a

> > Let the gap s^2 = c^2 - a^2 = (c - a)(c + a) = 2ab + b^2 which we

This is essentially an assumption that c-a < 1

which is an assumption that Andrica's conjecture is true.

Is this supposed to be an "assume for now" type statement as

if it will be proven correct or somehow justified later in the proof?

> > Take the square root to find c

This step above is a trouble spot,

> > c = a + b

> > s^2 = 2ab + b^2 < a + b + a = 2a + b

> > Or 2ab + b^2 < 2a + b

> > So 0 < b < 1

> > Replace b with y = 1/b so 1 < y however

> > 2a + y must be < a^2 by Bertrand/Tschebycheff:

how do you arrive at it?

By B/T the gap=2ab+b^2 < a^2=P(n)

since b=1/y, 2a/y + 1/y^2 < a^2

thus 2a+y < y*a^2-1/y+y

-[snip]- the rest.

The only way to prove Andrica's Conjecture

via Bertrand/Tschebycheff is to prove that

it is always true that

sqrt(2*p)-sqrt(p)<1

sqrt(2)-1<1/sqrt(p)

which clearly, is not true for all p>5.

You need stronger mojo baby.

-Dick