anonymous
  • anonymous
Hint for Number Theory homework 1) find the pattern of the following: a) (3^10) modulo 11 b) (2^12) modulo 13 c) (5^16) modulo 17 d) (3^22) modulo 23 2) Propose a theorem I found the answer: 1) It is all 1 2) The theorem is as following: (a^(n-1) modulo n) = 1; where n is prime; n >= 3 and a > 0 There is not proof required for homework; But I want to prove the theorem; but I am stuck since induction does not seem to work due to variance nature of n; any hint. Thanks. KingGeorge is needed. Thanks. Regards.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
it has possibly to do with relatively prime stuff
amistre64
  • amistre64
if p and q are prime, then p mod q = p p^(q-1) mod q p^(q) p^(-1) mod q p^q mod q = 1, proof it?
amistre64
  • amistre64
king george might be more elegant, i agree :)

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anonymous
  • anonymous
Hi Amistre64, Thanks a lot for your hints. In my proposed theorem, "a" does not have to be prime and it is still works. (9^16 modulo 17)=1. So does the hint still apply? I just started the course about 2 week ago and King George was the first to help. More elegance is more better. Thanks again Amistre64.
amistre64
  • amistre64
in the examples provided, they used primes. you will notice that 9 and 17 are relatively prime, meaning that their gcd is 1
anonymous
  • anonymous
Hi Amistre64, Maybe, "a" is prime will work for all cases. "a" including non-prime might work only occasionally. Hence my proposed theorm is wrong. Thanks.
amistre64
  • amistre64
4^7 mod 8 = 0
anonymous
  • anonymous
Hi Amistry64, It is a^(n-1) modulo n; where: n is prime. in your case 8 is not prime. I am missing something. Thanks again.
amistre64
  • amistre64
itll only work out if gcd(a,n) = 1 fermats little thrm is: a^(p-1) = 1 mod p, as long as p does not divide a theres proofs for that
anonymous
  • anonymous
Hi Amistry64, Thanks And I noticed the "a" to be prime in the sample; However, I checked out non-prime as well. So I made my proposed to be non-prime as well. It is risky; but more general; hence I need to prove before commit. Thanks.
anonymous
  • anonymous
Hi Amistry64, Thanks and great stuff; I am not aware of "little" theorem as yet; So I can proposed: gcd (a, p) = 1 for general case. Thanks.
amistre64
  • amistre64
consider a and n, such that they are relatviely prime: the set: a,2a,3a,4a,...,(n-1)a, none of which are divisible by n then the construct: a*2a*3a*4a*...*(n-1)a is equal to a^(n-1) a*2a*3a*4a*...*(n-1)a = 1*2*3*4*...*(n-1) (mod n) a^(n-1) = (n-1)! (mod n) it stands to reason that any multiple of a^(n-1) is also congruent (n-1)! mod n therefore a^(n-1) * (n-1)! = (n-1)! (mod n) since the gcd(n, (n-1)!) = 1 we can divide out the (n-1)! a^(n-1) = 1 (mod n)
amistre64
  • amistre64
this might only hold of n is prime
anonymous
  • anonymous
Hi Amistre64, Sorry for misspelling (Amistry64). :-) :-) :-) Thanks a lot for your answer; I will try not to see your answer; I need to try it for myself first; So I guess, the hints you gave is enough??? Thanks.
amistre64
  • amistre64
they are sufficient yes :) and good luck
anonymous
  • anonymous
Thanks again. Best regards.

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