Complete the following proof by contradiction

Question

Let a, b and c be integers. If 3 divides a, 3 divides b and $c \equiv 1(mod 3)$ then the equation $ax + by = c$ has no solution which x and y are both integers.

Complete the following proof by contradiction.

So we assume that the statement is false. That is, we assume that there exists integers a, b, and c such that 3 divides both a and b, and that $c \equiv 1(mod 3)$, and that the equation $ax + by = c$ has a solution in which x and y are both integers.

Now I know I can substitute a and b for $3k_1$ and $3k_2$ and c for $3k_3 + 1$ making the equation $3k_1x + 3k_2y = 3k_3+1$ but am completely stumped on where to go from here.

Answer 1

It follows from the equality $3k_1x + 3k_2y = 3k_3+1$ that $1=3(k_1x+k_2y-k_3)$, which is impossible, because $3$ does not divide $1$.

Answer 2

Note that $$ 3(k_1x + 3k_2y-k_3)=1 $$ Note that $3$ divides the LHS but not the RHS. Contradiction.