log(k*p +/- 1)/(p-1) = log(q)/2
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From: Gerry Myerson on 16 May 2010 20:23 In article <588745340.144280.1274020233739.JavaMail.root(a)gallium.mathforum.org>, "M.A.Fajjal" <h2maf(a)yahoo.com> wrote: > > 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 > > ...... Have you looked at pseudoprimes? Strong pseudoprimes? Carmichael numbers? -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: M.A.Fajjal on 16 May 2010 17:42 > In article > <588745340.144280.1274020233739.JavaMail.root(a)gallium. > mathforum.org>, > "M.A.Fajjal" <h2maf(a)yahoo.com> wrote: > > > > 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 > > > > ...... > > Have you looked at pseudoprimes? Strong pseudoprimes? > > Carmichael numbers? > pseudoprimes, Strong pseudoprimes and Carmichael numbers can pass for some q not all q example p := 18721:; n := (p-1)*(1/2):; is((2^n-1)/p, 'integer'); false is((3^n-1)/p, 'integer'); true is((5^n-1)/p, 'integer'); false p := 41041:; n := (p-1)*(1/2):; is((2^n-1)/p, 'integer'); true is((3^n-1)/p, 'integer'); true is((7^n-1)/p, 'integer'); false
From: Gerry Myerson on 17 May 2010 00:19
In article <252435540.165597.1274060570728.JavaMail.root(a)gallium.mathforum.org>, "M.A.Fajjal" <h2maf(a)yahoo.com> wrote: > > In article > > <588745340.144280.1274020233739.JavaMail.root(a)gallium. > > mathforum.org>, > > "M.A.Fajjal" <h2maf(a)yahoo.com> wrote: > > > > > > 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 > > > > > > ...... > > > > Have you looked at pseudoprimes? Strong pseudoprimes? > > > > Carmichael numbers? > > > pseudoprimes, Strong pseudoprimes and Carmichael numbers can pass for some q > not all q > > example > p := 18721:; > n := (p-1)*(1/2):; > > is((2^n-1)/p, 'integer'); false > is((3^n-1)/p, 'integer'); true > is((5^n-1)/p, 'integer'); false > > > p := 41041:; > n := (p-1)*(1/2):; > > is((2^n-1)/p, 'integer'); true > is((3^n-1)/p, 'integer'); true > is((7^n-1)/p, 'integer'); false Well, I suppose it's trivial that a composite number can't pass for all q; after all, if p = r s, where r is prime, then (r^n +/- 1) / p can't be an integer. This even shows up in your last example - 41041 = 7 x 5863. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email) |