## Proof of D. Andrica conjecture.

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• http://www.primepuzzles.net/conjectures/conj_008.htm Proof of D. Andrica conjecture. Two squares one with area P(n) the other P(n+1) where P(n+1)
Message 1 of 2 , Aug 2, 2002
http://www.primepuzzles.net/conjectures/conj_008.htm

Proof of D. Andrica conjecture.

Two squares one with area P(n) the other P(n+1) where P(n+1) < 2*P(n) by proof of "Bertrand's postulate" by Tschebycheff.

Subtract the area of the square P(n)/P(n) = 1: So the area
P(n+1)- P(n) = area D, the differance between the primes, is
(P(n+1)-P(n))/P(n) < 2-1 < 1.

So the sides have lengths of the squares are
sqrt(P(n+1)) < sqrt(2*p(n))and sqrt(P(n)) < sqrt(P(n+1). Or after dividing by sqrt(P(n))
1 < sqrt(P(n+1))/sqrt(P(n)) < sqrt(2)
0 < sqrt(P(n+1))/sqrt(P(n))- 1 < sqrt(2)-1 < 1.

The reason we can divide by sqrt(P(n)) is P(n+1) - P(n) = D is the differance between to squares.
D = (sqrt(P(n+1))- sqrt(P(n))) (sqrt(P(n+1))+sqrt(P(n)))
with
2*sqrt(P(n)) < sqrt(P(n+1))+sqrt(P(n))
making
D /sqrt(P(n))~(sqrt(P(n+1))-sqrt(P(n)))*(2 *sqrt(P(n)))/sqrt(P(n))
So QED

Do y'all See any problems?

[Non-text portions of this message have been removed]
• ... Maybe, Bertrand/Tschebycheff: pn
Message 2 of 2 , Aug 3, 2002
--- 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
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