Suppose that a, b, m, and n are integers (n, m positive) and suppose we know a = b (mod n) and n | m Can we safely conclude that a = b (mod m)? (a) True by transitivity (b) False because I can find small integers for which it fails. True, because congruence mod n has more equivalence classes so the conclusion is just a weaker (c) version of the hypothesis. True, because congruence mod n has fewer equivalence classes so the conclusion is just a weaker O (d) version of the hypothesis.

2. Prove that if a, b, c, d, and m are integers such that m > 0, a = b mod m and c= d mod m₂ then (a) a+c=b+d mod m (b) a-c=b-d mod m (c) ac = bd mod m

Apply properties of congruence to make computation in modulo n feasible. Find the unique r such that a is logically equivalent to r (mod n ) with 0 less than or equal to r and r less than n.  a) (25)2022(27)2022 ≡ r (mod 5) b)(1234567891237456789)2(1234567891)2022 ≡ r(mod 10) c) (3)2022 (9)2022 ≡ r (mod 27)

4. There are no integers a, b such that a² + b² = 3 mod 4.

Fermat's Little Theorem says that for all natural numbers p and integers x, if p is prime then xº = x mod p. (i) Show that the statement is wrong, if we drop the assumption that p is prime. [Remember: All you need to do is to give a counterexample.] (ii) Show, by giving an example, that if p is not prime, then xP = x mod p can still be true for some integers x with x -1, 0, 1.