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prime between n^2 and (n+1)^2

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  • gll
    I would like for you real mathematicians to tell me if this approach has any promise, or is it like so many others which seem plausible only to turn out to be
    Message 1 of 1 , Dec 1, 2002
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      I would like for you real mathematicians to tell me if this approach has any promise, or is it like so many others which seem plausible only to turn out to be undoable. I haven't seen any postings directly concerned with this approach except for my recent response to Dick Boland; so I don't know if there is anything new here or if this is just another trivial, elementary, but unfruitful concept.

      Notation for following (all variables represent integers): p_i^x1 is the ith prime number raised to the x1 power, p_n$ is any prime (not the nth) between n^2 and (n+1)^2, p_m is the largest prime < sqrt[(n+1)^2]=(n+1), abs(c) is the absolute value of c.

      We seek a prime number p_n$ between the squares of any two consecutive integers. We want n^2 < p_n$ < (n+1)^2. If we were to test an integer < (n+1)^2 to determine if it is prime, we would need to test divisibility of that number by all primes p_i <= p_m. So, any composite integer between n^2 and (n+1)^2 exclusively must have a factor p_i<= p_m < (n+1).

      We make use of this Theorem 1: For integers a, b, & c where a +/- b = c, it is impossible for exactly two of these integers to be divisible by any p_i. Proof: Assume to the contrary that there exists such integers a, b, & c where a +/- b = c, and that exactly two (say a and b) are divisible by some p_i. Then a = g*p_i and b = h*p_i where g and h are integers. Therefore, a +/- b = g*p_i +/- h*p_i = p_i*(g +/- h) = c. So c is divisible by p_i which is a contradiction.

      Now consider a +/- b = c where a = p_1^(x1)*p_2^(x2) * ... *p_m^(xm) where all xi are nonnegative integers and b = p_1^y1*p_2^y2*...*p_m^(xm) where all yi are nonnegative. Furthermore, let either xi = 0 or yi = 0 (but not both) for all i =1 to m so that either a or b but not both are divisible by each p_i, i=1 to m. Then by Theorem 1 c is not divisible by any p_i, i=1 to m. So, if abs(c) < (n+1)^2 then abs(c) is prime.

      Then to show there exists a prime number p_n$ between the squares of any two consecutive integers, we need to show that n^2 < abs(c) < (n+1)^2 where c is as defined above. Actually, the plus is applicable only for small n.

      So the problem reduces to the need to show that there exists n^2 < abs(a-b)= abs[p_1^(x1)*p_2^(x2) * ... *p_m^(xm) - p_1^y1*p_2^y2*...*p_m^(xm)] < (n+1)^2 for all n where p_m < (n+1). I think this may be doable but not by me.

      Here is an example. I seek a prime between n^2= 8^2=64 and (n+1)^2=9^2=81. So p_m=7and c =2^(x1)*3^(x2)*5^(x3)*7^(x4)-2^(y1)*3^(y2)*5^(y3)*7^(x4). Two solutions are c = 2^(2)*3*7-5=79 and 2^(3)*3*5-7^(2)=71.




      g long



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