(1) This is false. Let a=2, b=4, x=7, y=3, and d=26. It is clear that 2(7) + 4(3) = 26, but gcd(7,3) = 1 which is different from 26.
(2) This is true. Suppose that gcd(x,y) = k. Then, k | x and k | y, that is, x = sk and y = tk for some integers s and t. From ax + by =1, we have a(sk) + b(tk) = 1 yielding k(as + bt) =1 and so k | 1. This implies that k = 1 or -1. As k is the largest common divisor of x and y, then we must have k = 1, that is, gcd(x, y) = 1.
(3) Since a ≡ b (mod n) and x ≡ y (mod n), then a - b = kn and x - y = sn for some integers k and s, that is, a = kn + b and x = sn + y. It follows that ax = (kn + b)(sn + y) = ksn^2 + kyn + bsn + by = n(ksn + ky + bs) + by which yields ax - by = n(ksn + ky +bs), a multiple of n. This shows that ax ≡ by (mod n).
(4) Since gcd(a, b) =d, then there exist integers s and t such that as + bt =d. Dividing d on both sides yields (a/d)s + (b/d)t = 1, that is, a's + b't =1, where a' = a/d and b' = b/d. Using the result proven in (2), we see that
gcd(a', b')=1, that is, gcd(a/d, b/d)=1.
