Q:

How do you prove that $a + 1 = b$ in a field with 4 elements?

I have a Field with 4 distinct elements {$0, 1, a, b$}. And because $a \cdot b \neq a$, and vice versa, I have filled the $\cdot$ table as:
$$
        \begin{matrix}
        \mathbf\cdot & \mathbf0 & \mathbf1 & \mathbf{a} & \mathbf{b} \\
        \mathbf0 & 0 & 0 & 0 & 0 \\
        \mathbf1 & 0 & 1 & a & b \\
        \mathbf{a} & 0 & a & b & 1\\
        \mathbf{b} & 0 & b & 1 & a\\
        \end{matrix}
$$
Now to fill the $+$ table, I can't figure out how to prove by contradiction that $a + 1 = b$, and similarly, $b + 1 = a$.
Any pointers on how to proceed?
EDIT: I'm thinking that the distributive law fits in here somewhere?

A:

If $a+1=a$, then we have $1=0$, impossible.
If $a+1=1$, then we have $a=0$, impossible.
If $a+1=0$, we multiply by $a$ on both sides, getting $a^2+a=0$ or $b+a=0$.  So $b=1$, impossible.

