Prove if $a|3c-b$ and $6c \equiv 4b \pmod a$ then $a|2b$

Question

I had this question on a quiz, I didn't get it right. We got the quizzes back and I tried it again. Is this correct?

Question: Prove if $a|3c-b$ and $6c \equiv 4b \pmod a$ then $a|2b$

Since $a|3c-b$, $\exists k_1$ such that $ak_1 = 3c-b$

Since $6c \equiv 4b \pmod a$, $a|6c-4b$, $\exists k_2$ such that $ak_2 = 6c - 4b$

We want $a |2b$, meaning that $\exists k$ s.t. $ak = 2b$

This is where I went wrong in the quiz, but I have tried again. The TA only said Let $k = 2k_1 - k_2$:

$\begin{gather} ak_2 = 6c-4b \\ 2ak_1 = 6c-2b \\ \text{subtract}\\ 2ak_1 - ak_2 = -6b \\ a(2k_1 - k_2) = -6b \\ \end{gather}$

So let $k=2k_1 - k_2$

Then:

$\begin{gather} ak = 2b\\ a(2k_1 - k_2) = 2ak_1 - ak_2 = 2b\\ 2(3c-b) - (6c-4b) = 2b \\ 6c-2b - 6c + 4b = 2b \\ 2b = 2b \end{gather}$

I feel like I reached the right conclusion but the whole $a(2ka_1 - k_2) = -6b$ is...well I don't think I did that right.

Thanks

Answer 1

No, it is not correct,

$\begin{gather} ak_2 = 6c-4b \\ 2ak_1 = 6c-2b \\ \text{subtract}\\ 2\color{red}{ak_1} - ak_2 = \color{red}2b \\ a(2\color{red}{k_1}-k_2)=\color{red}2b \end{gather}$

Hence we can conclude that $a|2b$ since $2k_1-k_2 \in \mathbb{Z}$.