Thread: A tentative question View Single Post
2020-07-14, 12:27   #22
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by devarajkandadai This example is not correct.correct example: 23 is non-residue of 3571 upto infinite order.I leave it to pari experts like Charles to verify.
Hey clueless. You can't even do simple arithmetic. Nor do you bother to study any
math. The result makes you look stupid every time you open your mouth.

Consider any prime q that does NOT divide (3571-1). Take, e.g. q = 11

note that 1148^11 = 23 mod 3571.

23 is an 11'th order residue mod 3571. I think we all safely know (except perhaps you)
that 11 is finite. You should now be asking "what is special about 11?"

If you had bothered to take my earlier hint about cubic residues modulo a prime
that that is 1 mod 6 vs. primes that are -1 mod 6 you might have avoided this
latest erroneous assertion. I will give a further hint: Only 1/3 of the residues
less than p are cubic residues of p when p = 1 mod 6. But when q = -1 mod 6,
they ALL are. Learning WHY is directly tied into Lagrange's Theorem. It is also
tied into the Sylow theorems. [Ask yourself how many subgroups there are of size
(p-1)/3]

Go learn some mathematics. In particular learn Lagrange's Theorem. Learn
Euler's Theorem for quadratic reciprocity. Study its generalization. Learn what
a primitive root is. Read and study the Sylow theorems.

Consider the following:

Prove or disprove:

For prime p,q, x^q = a mod p always has a solution for every a when q does not divide p-1.

Then ask: What happens if q | (p-1)???

Go read and study Nick's excellent introduction [in this forum] to number theory.
I think we all know that you will ignore this advice.

Finally STFU until you can be bothered studying at least some of this subject.
If you want to post mindless numerology go to the misc.math sub-forum or
open your own sub-forum in the blogorrhea.