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.
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
it has possibly to do with relatively prime stuff
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?
Not the answer you are looking for? Search for more explanations.
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.
in the examples provided, they used primes.
you will notice that 9 and 17 are relatively prime, meaning that their gcd is 1
Maybe, "a" is prime will work for all cases. "a" including non-prime might work only
occasionally. Hence my proposed theorm is wrong. Thanks.
4^7 mod 8 = 0
It is a^(n-1) modulo n; where: n is prime. in your case 8 is not prime. I am missing
something. Thanks again.
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
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.
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.
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
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)
this might only hold of n is prime
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.