Writing a proposition as the conjunction of two conditional statements For each integer a, a is congruent to 3(mod 7) if and only if (a^2 + 5) is congruent to 3(mod 7). 
A)Writing a proposition as the conjunction of two conditional statements 
B) determine if the conditionals are true or false 
 A: For completeness and because any deadline has probably passed for this homework-type quesiton, I will construct a full solution.
A) Let $A_a$ be defined as "$a$ is congruent to $3\bmod 7$" and $B_a$ defined as "$a^2+5$ is congruent to $3\bmod 7$." We are given $(\forall a \in \mathbb{Z})A_a \equiv B_a$, which by definition is composed of two conditionals $(\forall a \in \mathbb{Z})(A_a \rightarrow B_a) \wedge (B_a \rightarrow A_a)$ .
So, the full answer should be: For all integers $a$, (If $a$ is congruent to $3 \bmod 7$ then $a^2 + 5$ is congruent to $3 \bmod 7$) and (if $a^2 + 5$ is congruent to $3 \bmod 7$ then $a$ is congruent to $3 \bmod 7$).
B) 
(1)Start with the first conditonal, $(\forall a)(A_a \rightarrow B_a)$: If $a$ is congruent to $3 \bmod 7$ then $a^2 + 5$ is congruent to $3 \bmod 7$. Suppose that ($\neg A$) $a$ is NOT congruent to $3 \bmod 7$ then the conditional is vacuously true, so we don't need to further check this case. Suppose ($A$ is true) $a$ is congruent to $3 \bmod 7$ this means that $7 \lvert a-3 $. We need to check that ($B$) $7\lvert ((a^2+5)-3)$.
$$\frac{a^2 + 5 - 3}{7} = \frac{a^2 +2}{7} = \frac{(a-3+3)(a-3+3)+2}{7} = \frac{(a-3)^2+6(a-3)+11}{7}$$The first two terms are clearly divisible by $7$ given that $7\vert a-3$ but the third is not. So, the consequent of the conditional is false and thus $(\forall a)(A_a \rightarrow B_a)$ is false.
(2) The second conditional says $(\forall a)(B_a \rightarrow A_a)$. As before if $B$ is false, then the conditional is vacuously true, but we must check what happens when $B$ is true. Suppose $7\vert (a^2 +2)$, then $a^2 \equiv -2 \ (\bmod 7) \Rightarrow a^2 \equiv 5 \ (\bmod 7)$ but the squares in the integers $ \mod 7$ are just $\{1,2,4\}$ so this equation has no solutions. The antecedent is then always false, so the conditional, $(\forall a) (B_a \rightarrow A_a)$ is true. 
