coldize

Hey I'm trying to prove a biconditional statement: Let a be an integer. a is congruent to 2 modulo 5 if and only if a squared is congruent to 4 modulo 5. I proved it to the right and and I'm working on proving it to the left. I thought a proof by contradiction would be best but what I run into is the equation 5(b/a+2) = a-2 for some integer b. Now, if I could somehow prove that (b/a+2) is an integer, I would be set because that would contradict our assumption that 5 does not divide a-2 (because we're doing a proof by contradiction). So if anyone could help or point me in the right direction, that would be wonderful. I was hoping these forums would have LaTeX embedding capabilities but it appears that isn't possible so I've just attached the LaTeX code I do have as a code snippet. Thanks again! \begin{proof}[Part 2] We will prove that if $a^2 \equiv 4$ (mod 5), then $a \equiv 2$ (mod 5) by using a proof by contradiction. That is, we will prove that if $a^2 \equiv 4$ (mod 5)and $a \not \equiv 2$ (mod 5), then there will exist a logical fallacy.\\ \\Assume that $a^2 \equiv 4$ (mod 5) and $a \not \equiv 2$ (mod 5). By the definition of congruency, we know that $5|a^2-4$ and $5 \nmid a-2$. By the definition of divides, we know that $\exists b \in \mathbb{Z}$ such that $5b=a^2-4$. Observe that $a^2-4 = (a+2)(a-2)$. By dividing both sides of the equation $5b=(a+2)(a-2)$ by $(a+2)$, we obtain the equation $5(b/a+2)=(a-2)$. Since $b/a+2$ can be rewritten ..... ? \end{proof}