Recent content by Ch1ronTL34

Find the range and kernel of: a) T(v1,v2) = (v2, v1) b) T(v1,v2,v3) = (v1,v2) c) T(v1,v2) = (0,0) d) T(v1,v2) = (v1, v1) Unfortunately the book I'm using (Strang, 4th edition) doesn't even mention these terms and my professor isn't helpful. My professor said: "Since range and kernel...

The question is: Show that if p is an odd prime and ord(p^a)a=2t, then a^t== -1 mod p^a First, I used ord(p^a)a to mean "order of a, mod p^a" and the == sign means congruent. So first, I tried a few examples. Let p=3, a=2 Since ord(9)2=6, then t=3 and: 2^3 == -1 mod 9 TRUE I...

Ok, my question is: Show that if ab == 1 mod m, then ord(m)a=ord(m)b (Note that == means congruent) and ord(m)a means the order of a mod m I know that if a^k==1 mod m, then the ord(m)a is the smallest integer k such that the congruence holds. For example, ord(10)7=4 since 7^4==1 mod...

Alright I think I get it now. Thanks so much for the help shmoe!

Yes shmoe, I do have a textbook..."Elementary Theory of Numbers" by William J. LeVeque. Unlike most mathbooks, this one is about 100 pages and very small...though its size has proven to be misleading...ANYWAYS: When p==1 mod 4, (-1)^(p-1)/2 will be an even number because p will be odd (p=4d+1...

This is for a class but he didn't teach us ANYTHING about the legendre symbol. I'm guessing that 1 mod 4 implies (-1/p)=1 is critical but i still don't really understand. I took shmoe's advice and found that, when p==1 mod 4 a^(p-1)/2 mod 5 equals (-a)^(p-1)/2 mod 5

No shmoe, I don't know anything about (-1/p) and I just did some research on Euler's criterion: if a is quadratic residue modulo p (i.e. there exists a number k such that k^2 ≡ a (mod p)), then a^(p − 1)/2 ≡ 1 (mod p) and if a is not a quadratic residue modulo p then a^(p − 1)/2 ≡ −1...

The question is: Show that if p == 1 mod 4, then (a/p) = (-a/p). (Note that == means congruent). I know that if X^2==a mod p (p is a prime) is solvable then a is a quadratic residue of p. For an example, I let p = 5 since 5==1 mod 4. Then, I let X = 2 and 4 just to check the equation...

ok i wrote out the prime factorization of the equation in the first post. The d on the numerator cancels the d in the denominator. I'm not very good thinking about primes and factors (even though I am taking Number theory haha). Which prime factors does d have? Does d have the same...

I don't exactly understand what you mean by "which, remember will be in d" d is the gcd of a and b

Status X...that was another part of the question. They gave us the fact that phi(a)=a*PROD(1-1/p) for the distinct prime factors. Can one say: phi(ab)=ab*PROD(1-1/p) for the distinct prime factors of a and b? I don't know if this would help me or not

Ok the question is as follows: Given gcd(a,b)=d, show that Phi(ab)= (d*phi(a)phi(b))/phi(d) I know that if gcd(a,b)=1 then phi(ab)=Phi(a)phi(b) but I am just stuck here. Any help would be greatly appreciated!