1. The problem statement, all variables and given/known data 3.) For each of the following a mod n, find a≡1 mod n, i.e. find an integer x with ax≡1 mod n, or explain why such an integer does not exist: 5' mod 11; (21)' mod 28; 2' mod 101; 4' mod 101: 4.) Let n = 2k+1 be an odd number. What is 2'≡1 mod n? What happens when n = 2k is even? Now find the multiplicative inverse of 3 mod n: First suppose n = 3k + 2 for some k. What is 3' mod n? (It looks a lot like the answer for 2'.) If n = 3k + 1, find 3' mod n as follows. First find a number x such that 3x≡-1 mod n, and then note that 3(-x)≡1 mod n; finally, write -x as a number between 0 and n - 1. What if n = 3k? 2. Relevant equations ax+by=c, has a solution if gcd(a,b)|c Euclidean Algorithm 3. The attempt at a solution So I'm just not sure I've done these problems the right way, or the most efficient way. My answers are often negative, but I figured since its congruence classes, when my answers are -x, it really means [-x] which should be okay? But is that a cop-out? Should I be trying to find representatives between zero and n? Question 4 seems to imply that, but I didn't exactly do the problem the way my teacher suggested, I did it how it made sense to me. So could someone check my work? Also, in Algebra in general I feel like I write my answers in a weird way or in an odd order. If it seems I've done that, please let me know how you would write your answers in a more elegant or simple to understand way. 3) ax≡1 mod n → ax+kn=1 gcd(5,11)= 1 = 11-2(5) → 5' mod 11= -k gcd(21,28)= 7 = 28-21. 7 does not dive 1, therefore no solution exists. gcd(2,101)= 1 = 101-50(2) → 2' mod 101= -50 gcd(4,101)= 1 = 101-25(4) → 4' mod 101= -25 4)n=2k +1, n is odd, 2': gcd(2,2k+1)=1= (2k+1)+ (-k)(2) → 2' mod (2k+1)= -k n=2k 2': gcd(2,2k)= 2. 2 does not divide 1→ solution does not exist. n=3k+2 3': gcd(3,3k+2)= 1= 2(3k+2)+ (-2k-1)(3) → 3' mod (3k+2)= -2k-1 n=3k+1 3': gcd(3,3k+1)=1 = (3k+1)+ (-k)(3) → 3' mod (3k+1)= -k n=3k 3': gcd(3,3k)= 3. 3 does not divide 1 → solution does not exist.