log(k*p +/- 1)/(p-1) = log(q)/2
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From: M.A.Fajjal on 16 May 2010 07:10 > M.A.Fajjal : > > > > > > > > > p is a prime iff there is an integer k such > > that > > > > > > > > > > log(k*p +/- 1)/(p-1) = log(q)/2 > > > > > > > > > > where q is any prime > > > > > > > > > > Is there any counter example > > > > > > > > q=/=p > > > > > > reduce to log( (k*p +/- 1)^(p-1) ) = log(q^2) > > ?? > > It should be > > log( (k*p +/- 1)^(1/(p-1)) ) = log(q^(1/2)) > > > > > > > reduce to (k*p +/- 1) ^ ((p-1)/2) = q > > > > > It should be > > > > (k*p +/- 1) ^ (2/(p-1)) = q > > yes sorry. > > but still , > for integer A and rational B -> A^B =/= prime ! > but for a prime p it will be examples k=89 ; p=23 then (k*p + 1) ^ (2/(p-1)) = 2 k=164930; p=29 then (k*p - 1) ^ (2/(p-1)) = 3
From: master1729 on 16 May 2010 08:23 M.A.Fajjal : > > > > > > > > > > > > p is a prime iff there is an integer k > such > > > that > > > > > > > > > > > > log(k*p +/- 1)/(p-1) = log(q)/2 > > > > > > > > > > > > where q is any prime > > > > > > > > > > > > Is there any counter example > > > > > > > > > > q=/=p > > > > > > > > reduce to log( (k*p +/- 1)^(p-1) ) = log(q^2) > > > ?? > > > It should be > > > log( (k*p +/- 1)^(1/(p-1)) ) = log(q^(1/2)) > > > > > > > > > > reduce to (k*p +/- 1) ^ ((p-1)/2) = q > > > > > > > It should be > > > > > > (k*p +/- 1) ^ (2/(p-1)) = q > > > > yes sorry. > > > > but still , > > for integer A and rational B -> A^B =/= prime ! > > > but for a prime p it will be > > examples > > k=89 ; p=23 then (k*p + 1) ^ (2/(p-1)) = 2 > > k=164930; p=29 then (k*p - 1) ^ (2/(p-1)) = 3 ah i see what you mean now. perhaps intresting ... is there a simple way to find the smallest k when given p ? are you sure this is a new idea ? regards tommy1729
From: master1729 on 16 May 2010 08:39 > > > > > but for a prime p it will be > > > > examples > > > > k=89 ; p=23 then (k*p + 1) ^ (2/(p-1)) = 2 > > > > k=164930; p=29 then (k*p - 1) ^ (2/(p-1)) = 3 > > ah i see what you mean now. > > perhaps intresting ... > > is there a simple way to find the smallest k when > given p ? > > are you sure this is a new idea ? > > regards > > tommy1729 i guess you got inspired by fermat's little ... i wonder what else. i notice you wrote iff and not if. i dont yet see why it fails for all composite p ... regards tommy1729
From: M.A.Fajjal on 16 May 2010 09:42 > > > for integer A and rational B -> A^B =/= prime ! > > > > > but for a prime p it will be > > > > examples > > > > k=89 ; p=23 then (k*p + 1) ^ (2/(p-1)) = 2 > > > > k=164930; p=29 then (k*p - 1) ^ (2/(p-1)) = 3 > > ah i see what you mean now. > > perhaps intresting ... > > is there a simple way to find the smallest k when > given p ? q is given as any prime say 43 then for any p say 11 k = (43^5+1)/11 = 13364404 Also for q = 11 and p=43 there is another k k = (11^21-1)/43 = 172098835912980467470 for any integer say 15 k = (43^7-1)/15 = 90606203702/5 not integer k = (43^7+1)/15 = 271818611108/15 not integer > are you sure this is a new idea ? Not sure
From: M.A.Fajjal on 16 May 2010 09:48
> > > > > > > but for a prime p it will be > > > > > > examples > > > > > > k=89 ; p=23 then (k*p + 1) ^ (2/(p-1)) = 2 > > > > > > k=164930; p=29 then (k*p - 1) ^ (2/(p-1)) = 3 > > > > ah i see what you mean now. > > > > perhaps intresting ... > > > > is there a simple way to find the smallest k when > > given p ? > > > > are you sure this is a new idea ? > > > > regards > > > > tommy1729 > > i guess you got inspired by fermat's little ... > > i wonder what else. > > i notice you wrote iff and not if. If q is a given prime then for any prime p there exit an integer k such that q^((p-1)/2)+/-1 = k Also if there is exist an integer k such that q^((p-1)/2)+/-1 = k then p is prime > > i dont yet see why it fails for all composite p ... > Is there any example |