If a | b and b | c, then a^2 does not necessarily divide c, as shown by the counterexample where a=b=c=2. If a | b, then a | b^2, which is proven. While a | (b + c) does not imply that a | b and a | c, as demonstrated by the counterexample where a = 2, b = 1, c = 3.

1. Throughout, a, b, c epsilon Z. The following statement is false; give a counterexample: If a | b
and b | c, then a2 | c. Prove: If a | b then a | b2. The following statement is false; give a
counterexample: If a | (b + c), then a | b and a | c.
Solution
1. Let a=b=c=2. Clearly a|b and b|c but a^2 does not divide c. 2. Assume a|b. Then
for some integer m, am = b. So (am)^2 = b^2 => (am)(am) = b^2 => a(am^2) = b^2 Since am^2
is an integer, then a|b^2. 3. Let a = 2, b = 1, and c = 3. Clearly a|(b+c) but neither a divides b nor
a divides c.