log(k*p +/- 1)/(p-1) = log(q)/2
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From: M.A.Fajjal on 16 May 2010 02:46 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
From: M.A.Fajjal on 16 May 2010 02:54 > 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
From: master1729 on 16 May 2010 03:17 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) reduce to (k*p +/- 1) ^ ((p-1)/2) = q a^b cannot equal a prime q for a and b > 2. QED regards tommy1729
From: M.A.Fajjal on 16 May 2010 06:29 > 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 Note that q is given as any prime =/= p Say q = 2 Then for any prime p there exist integer k such that 2^((p-1)/2) +/- 1 = k*p in other words for any prime p modp(2^((p-1)/2) +/- 1,p) = 0 modp(3^((p-1)/2) +/- 1,p) = 0 modp(5^((p-1)/2) +/- 1,p) = 0 ...... modp(719^((p-1)/2) +/- 1,p) = 0 .......
From: master1729 on 16 May 2010 06:42
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 perhaps you mean some kind of modular aritmetic ?? |