primenumbers at Yahoo! Groups

Carmichael result

Sebastian Martin Ruiz — Sun, 05 Jul 2009 15:07:27 GMT

Hello: We define the Carmichael function as the smallest integer Lambda(n) such that k**Lambda(n)=1 (mod n) for all k relatively prime to n. I have

Re: Small prime divisors of very large numbers

David Broadhurst — Sun, 05 Jul 2009 14:14:11 GMT

... For information, here is the GP version that I used: GP/PARI CALCULATOR Version 2.4.2 (development) amd64 running linux (x86-64 kernel) 64-bit version

Re: Small prime factors of large integers: A closer look.

Maximilian Hasler — Sun, 05 Jul 2009 13:10:01 GMT

... Aloha ! Before publishing the proof, consider f(x) = a^x + c = 2^2 + 4 = 8 eulerphi(8) = 4 2^(2+k*4) + 4 = 4 (mod 8) for all k > 0. Maximilian

Re: Small prime divisors of very large numbers

David Broadhurst — Sun, 05 Jul 2009 11:59:17 GMT

... {mypmod(b,n,p)=local(m=[p],f=0); while(n=n-1,m=concat(eulerphi(m[1]),m)); for(p=n=1,#m,n=lift(Mod(b,m[p])^n);

Re: Small prime factors of large integers: A closer look.

Devaraj Kandadai — Sun, 05 Jul 2009 05:12:37 GMT

In other words if a^b^d...^x + c = f(x) (where ab,c...x belong to N, x is the only variable) Then a ^b^d...^(x +

Re: Small prime factors of large integers: A closer look.

David Broadhurst — Sun, 05 Jul 2009 00:22:32 GMT

... You proof is also good for composite p, provided that p < q and q is prime, as you were careful to stipulate. But what if p > q ? In that case iteration of

Re: Small prime divisors of very large numbers

David Broadhurst — Sat, 04 Jul 2009 23:15:57 GMT

... The right answer by a bogus method :-) My mathematically legal method involves 3 applications of the CRT, in this "znorder" chain: z(x)=znorder(Mod(137,x))

Re: Oldie but a goodie...

Phil Carmody — Sat, 04 Jul 2009 17:32:26 GMT

... Ah - memories of http://www.leyland.vispa.com/numth/primes/xyyx.htm Phil

Re: Small prime divisors of very large numbers

richard_in_reading — Sat, 04 Jul 2009 16:58:32 GMT

... Ok. I'm probably going to blot my copybook here but I think the last digits of 137^^k mod p are k=1 0000000137 k=2 7801462217 k=3 0547325344 k=4 5375909947

Small prime factors of large integers: A closer look.

Kermit Rose — Sat, 04 Jul 2009 13:52:30 GMT

Small prime factors of large integers: A closer look. 137**(137**(137**137) mod 29 EulerPhi(29)=28 137**(137**137) mod 28 EulerPhi(28) = 12 137**137 mod 12

Re: Small prime divisors of very large numbers

David Broadhurst — Sat, 04 Jul 2009 10:15:24 GMT

... Simply ask Pari-GP for Mod(137,1+808*137^138)^(137^137) Puzzle: Find last 10 decimal digits of the residue of

Re: Small prime divisors of very large numbers

Patrick Capelle — Sat, 04 Jul 2009 07:42:49 GMT

... Note that 73 and 137 are the smallest prime factors of 10^(8n-4) +1. They are also prime factors of 10^8n -1, where n is a positive integer. Useful?

Re: Small prime divisors of very large numbers

richard_in_reading — Sat, 04 Jul 2009 05:48:23 GMT

... Isn't it irritating to work out the right way of doing something and not being able to show the wrong way failing for the base of interest! What's

Primes of the form a^2-b^2

sopadeajo2001 — Sat, 04 Jul 2009 03:41:37 GMT

Primes can be represented as a^2-b^2=(a+b)(a-b) if we choose a-b=1 and a+b=p so a=(p+1)/2 and b=(p-1)/2 so a and b are consecutive integers one odd and another

Re: Small prime divisors of very large numbers

David Broadhurst — Sat, 04 Jul 2009 00:49:07 GMT

... The next step is to look at cases with a Hardy-Wright "gcd problem" that cannot be solved so easily and then solve them with the Chinese Remainder Theorem