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Topic: Request for help with short argument
Replies: 8   Last Post: Nov 15, 2012 2:52 PM

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trj

Posts: 91
From: Korea
Registered: 2/23/12
Re: Request for help with short argument
Posted: Nov 14, 2012 6:56 PM
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> argument
>
>
> To search for numbers n such that n-1 and n+1 are
> both prime, an initially appealing naive approach
> suggests n is chosen so that primes 2, 3, 5, 7, etc
> all divide into n. However n grows so quickly that at
> larger values of n there are soon composites of
> primes too large to divide n which are in the
> neighbourhood of n.
>
> With that approach every 30th number is an n dividing
> by 2, 3 and 5. However, instead note that of the five
> natural numbers n-2, n-1, n, n+1, and n+2, three of
> these, n-2, n, and n+2, might divide by 5 and still
> leave n-1 and n+1 not divisible by 5.
>
> This generalises to ((p_n) - 2)/(p_n) many n having
> the nearest multiple of p_n not equal to either n-1
> or n+1.
>
> By which the integer value of 1/2 . 1/3 . 3.5 . 5/7 .
> 9/11 ... ((p_n) - 2)/(p_n) gives a lower bound for
> the number of n in some given range where neither n-1
> nor n+1 divide by any of 2, 3, 5, 7, 11 .... p_n.
>
> Trying this approach we note that between the squares
> of two consecutive primes (p_n-1 squared and p_n
> squared) three numbers are immediately unavailable as
> candidates for this type of n (namely p_n-1 squared +
> 1 ; p_n squared - 1 ; and one other ..... for example
> there are 25 - 9 - 3 = 13 natural numbers between 25
> and 9 adjacent to neither 25 or 9, these being 11,
> 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23).
>
> We use the fact that if we can sieve out all
> composites below (p_n)^2 dividing by any prime up to
> p_n, then only primes remain.
>
> This gives us the following expression (1)
>
> (1) number of valid n between (p_n-1)^2 and (p_n)^2
> = integer value of
> 1/2 . 1/3 . 3/5 ..... ((p_n) - 2)/(p_n) times
> [(p_n)^2 - (p_n-1)^2 - 3],


Here you assume that divisibility by different primes p' and p"
are sufficiently independent of each other. What I mean is that when you
multiply by 3/5, say, then it is because you know that 3/5 of all numbers
in the entire range between (p_n-1)^2 and (p_n)^2 have the property
that both their neighbour numbers are not divisible by 5.
But the problem is that this is not what you want to know.
You want to know whether it is also true of those 1/6 of the
numbers in the same range that remain valid because neither 2 nor 3
divide their neighbours. Even though 3/5 of all numbers are good,
maybe you miss the 1/6 of them that you are really interested in.

> which, on multiplying the expression to the left of
> 'times' by (p_n) and dividing the term on the right
> in square brackets by (p_n), becomes expression (2)
>
> (2) number of valid n in that range = integer value
> of
> 1/2 (1/3 . 3/5 .... ((p_n-1) - 2)(p_n-1) . ((p_n) -
> 2) ..) times [(p_n) - ((p_n-1)^2) + 3)/(p_n)],
>
> and rearranging numerators to the left of 'times' (by
> moving each numerator inside the curved brackets one
> place to the left) to give expression (3)
>
> (3) number of valid n in that range = integer value
> of
> 1/2 (3/3.5/5.9/7 .... ((p_n) - 2)/(p_n-1) ..) times
> [(p_n) - ((p_n-1)^2) + 3)/(p_n)].
>
> The term in the square brackets to the right of
> 'times' in expression (3) gives an integer value of
> at least 3 (for example the three pairs of
> consecutive primes 5 & 7 ; 11 & 13 ; 17 & 19 yield
> values of 3 ; 3 + 6/13 ; 3 + 12/19 respectively).
>
> Since the left side multiplying the square brackets
> after the first couple of terms always > 1/2, the
> final value is the integer value of at least half 3,
> which is 1. Therefore there is at least one such n
> between the squares of any two consecutive primes,
> therefore infinitely many twin primes, p_B = p_A+2.
>
> // Anything salvageable from this?




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