 mersenneforum.org Proth theorem extended
 Register FAQ Search Today's Posts Mark Forums Read 2009-04-06, 14:45 #1 Bill Bouris   Apr 2009 2 Posts Proth theorem extended Proth theorem extended: let Q= k*2^n +1, where n=>3 is a odd natural number & k<= 2^n +1. if for some 'a', a^((Q-1)/4) == +/-1(mod Q), then 'Q' is prime. 'k' doesn't need to be restricted to only 'odd' numbers, either. proof: if 'm' is from the set of natural numbers, then every odd prime divisor 'q' of a^(2^(m+1))+/-1 implies that q == +/-1(mod a^(m+2)) [concluded from generalized 'Fermat-number' proofs by Proth along with my replacing 'm' with 'm+1']. now, if 'p' is any prime divisor of 'R', then a^((Q-1)/4) = (a^k)^(2^(n-2)) == +/-1(mod p) implies that p == +/-1 (mod 2^n). thus, if 'R' is compo-site, 'R' will be the product of at least two primes each of which has minimum value of (2^n +1), and it follows that... k*2^n +1 >= (2^n +1)*(2^n +1) = (2^n)*(2^n) + 2*(2^n) +1; but the 1's cancel, so k*(2^n) >= (2^n)*(2^n) + 2*(2^n) and upon dividing by 2^n... k>= 2^n +2. moreover, this result is incompatible with our definition, so if k<= 2^n +1 and a^((Q-1)/4) == +/-1(mod Q), then 'Q' is prime. finally, if either k= 2*m or k= 2*m +1, then 'Q' is still 'odd'; Q= 2*m*2^n +1= m*2^(n+1)+1, or Q= (2m+1)*2^n +1= m*2^(n+1) +2^n+1. *QED   2009-04-06, 16:44   #2
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

7×191 Posts Quote:
 Originally Posted by Bill Bouris Proth theorem extended: let Q= k*2^n +1, where n=>3 is a odd natural number & k<= 2^n +1. if for some 'a', a^((Q-1)/4) == +/-1(mod Q), then 'Q' is prime. 'k' doesn't need to be restricted to only 'odd' numbers, either. proof:
When I see such a "theorem" I try to find a counter-example in Pari-Gp. Here it is one:

Q=8355841, so n=15, k=255, and let a=3, the conditions are true.
a^((Q-1)/4)==1 mod Q, but Q is composite: Q=13*41*61*257   2009-04-07, 06:56 #3 10metreh   Nov 2008 2·33·43 Posts I vote Misc. Math.   2009-04-07, 07:46 #4 retina Undefined   "The unspeakable one" Jun 2006 My evil lair 3·1,811 Posts So where is the error in the "proof"? I think it would be quite instructive to see where it goes wrong.   2009-04-07, 13:25   #5
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

7×191 Posts Quote:
 Originally Posted by retina So where is the error in the "proof"? I think it would be quite instructive to see where it goes wrong.
Yes, this is far from a standard proof in math. Here it is a big mistake:
"now, if 'p' is any prime divisor of 'R', then a^((Q-1)/4) = (a^k)^(2^(n-2)) == +/-1(mod p) implies that p == +/-1 (mod 2^n)"
This is totally false.
And R=Q in the "proof", if you haven't observed it.

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