Final answer:
To solve the mathematics problems, we showed that if a divides both b and c, then a^2 divides b · c; identified six integers within the equivalence class of 12 modulo 7, including both positive and negative numbers; and calculated 128 mod 7 using equivalence classes to find the remainder is 3.
Step-by-step explanation:
Regarding the mathematics problems at hand:
(a) Proving that a^2 divides b · c
If a divides b and a divides c, there exist integers such that b = a · x and c = a · y. Multiplying both equations we get b · c = a^2 · x · y, proving that a^2 divides b · c.
(c) Positive and Negative Integers in Equivalence Class
For the equivalence class [12], given m = 7, positive integers can be 12, 19, 26 and negative integers can be 5, -2, -9 as each differs from 12 by 7n for some n.
(d) Finding 128 mod 7 Using Equivalence Classes
128 is equivalent to 3 modulo 7 as 128 = 7 · 18 + 3 and hence 128 mod 7 = 3.