Abstract algebra proof help please?

Question

In order to solve x^i ≡ a(mod m), one must find integers b and c satisfying ib - φ(m)c = 1. The solution is x ≡ a^i (mod m).

How does one prove that x ≡ a^i (mod m) is a solution to x^i ≡ a (mod m) even if the greatest common divisor of a and m is greater than 1?

Answer 1

if a = 2, and m =4,
then we have x ≡ 2^i mod4 and x^i ≡ 2 mod 4
if i =1 then it is true,
if i =2, then 2^2 =4 ≡ 0 mod 4
then x ≡ 0 mod 4.
on the other hand x^2 ≡ 2 mod 4, which does not have x=0 as one of the solutions.
so what you want to prove is NOT true

Answer 2

properly, the way i might initiate this evidence may well be to, on the outset, state the products you need to use. So... enable G be a team, |G| > a million, and enable H < G. anticipate |G| is countless. Then, |H| may well be something, which leads to a contradiction. So, |G| < inf. for that reason, by utilising Lagrange, |H| divides |G| on account that G is a finite team. enable |G| = n. for that reason, |H| = a million, n, or d, the place d is a divisor of n, d == a million or n. Case a million: |H| = a million. Then H = {e}. Case 2: |H| = n. Then H = G. Case 3: |H| = d. Then H is a suited subgroup of G (contradiction). for that reason, |H| = a million, n, ===> the sole divisors of n are a million and n, ===>n is a chief style. So, |G| is fundamental. --------------------------------- Following the define your professor gave you, the way you may use |g| < inf. is by utilising declaring that g^ok = e. So, ok divides |G|. So, |G| = km, m is an integer. So, |G| is finite. If each component had countless order, then there may well be no identitity to the gang that's a contradiction.