--- 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.

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

assume is < c + a

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

> > 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:

This step above is a trouble spot,

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