Question about linear congruences

Question

Consider the congruence

$$2x+7y \equiv 5\pmod{12}$$

Here $(2,7,12)=1$. Since $(2,12)=2$, we must have $$7y \equiv 5\pmod{2}$$

Which clearly gives $y \equiv 1\pmod{2}$, or $y \equiv 1,3,5,7,9,11\pmod{12}$

Why does the previous statement follow?. This is not a problem. It's something I can't understand from the chapter.

Answer 1

Suppose $[7y]_2 = [5]_2$, then $[7]_2 [y]_2 = [5]_2$, which implies that $[1]_2 [y]_2 = [1]_2$, that is, $y \equiv 1 (\mod 2)$. This implies that $y$ has to be odd.

Answer 2

$2x + 7y$ must be odd (since $5\mod 12$ is odd). But $2x$ cannot be odd, and therefore $7y$ must be odd. That means $y$ must be odd. What you've written is saying the same thing, only with modulo $2$ instead of even / odd.

Answer 3

The statement $2x + 7y \equiv 5 \pmod{12}$ means that $$2x + 7y = 5 + 12k$$ for some $k \in \mathbb{Z}$. We can consider this equation modulo $2$ to get $7y \equiv 5 \pmod{2}$, since both $2x$ and $12k$ are divisible by $2$. Because $7 \equiv 5 \equiv 1 \pmod{2}$ we get $y \equiv 1 \pmod{2}$, which means that we can write $$y = 1 + 2m$$ for some $m \in \mathbb{Z}$. Now you can consider this equation modulo $12$ to get $y \equiv 1 + 2m \pmod{12}$, by trying values for $m$ you can quickly see that $y$ can only be equivalent to $1,3,5,7,9$ or $11$ modulo $12$.

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It might give a little insight to look at the points in $\mathbb{Z}^2$ that correspond to the solutions of the original equation $2x + 7y \equiv 5 \pmod{12}$ shown below. It turns out that this this equation can be rewritten as $y \equiv 11-2x \pmod{12}$. We see a pattern of diagonal lines of dots appear. If you want to move around on a single diagonal line for every step you take in the $x$ direction you must take two steps in the $y$ direction because of the factor $2x$. This means that on a diagonal line the parity of the $y$ values can not change. But if you jump from one diagonal line up or down to a different diagonal line your $y$ value wil always change a multiple of $12$. Since twelve is even, this will also not change the parity of $y$. So we see that the $2x$ and the fact that the equation is in $\mathbb{Z}/12\mathbb{Z}$ work together to ensure that $y$ is odd for all solutions to the equation.