2.9.5

a) Consider the congruence $$2x \equiv 5 \ mod\ 9 $$; we claim that x is an integer that solves this congruence relation if and only if x has the form x = 9z + 7 for some integer z.

Observe that 5 is a multiplicative inverse of 2 modulo 9, for $$ 5*2 \equiv 10 \equiv 1\ mod\ 9$$. Hence, $$5*2x \equiv 25 \equiv 7\ mod\ 9 $$, but also $$5*2x \equiv (5*2)x \equiv x\ mod\ 9 $$, thus $$ x \equiv 7\ mod\ 9$$. So if x solves the congruence relation, x = 9z + 7 for some integer z. Conversely if x = 9z + 7 for some integer z, then 2x = 18z + 14 = 9(2z+1) + 5, so 2x is congruent to 5 mod 9.

b) Consider the congruence $$2x \equiv 5\ mod\ 6 $$; we claim there are no integers x which satisfy this congruence relation. Suppose for contradiction that such an x existed, then 2x = 6z + 5 for some integer z. 6z is even, thus 6z + 5 is odd. But if x is an integer, then 2x is even, so we have a contradiction.