Inverse function mod

Question

$$ y = 3x + 7 \pmod 4,\quad x,y \in \mathbb{Z}_4 $$

$x = 2, y =$ ??

Find the inverse function and verify the value of $y$.

I solved only the first question $y = 13\bmod 4 = 1 \bmod 4.

How to solve the second question for inverse?

Answer 1

Solve it like this (adding mod 4 after everything of course)

$$ y = 3x + 7\\ y - 7 = 3x\\ 3^{-1}(y-7) = 3^{-1}3x\\ 3^{-1}(y-7) = x $$

So all you need to do is find the inverse of $3$ mod $4$. Did you go over modular inverses yet?

Answer 2

The only possibilities are: $$(x,y) \in \{ (0,3),(1,2),(2,1),(3,0) \}$$ So the functions are as follows: $$y=\left\{\begin{array}{c} 3 & x=0 \\ 2 & x=1 \\ 1 & x=2 \\ 0 & x=3 \\ \end{array}\right\}$$ $$x=\left\{\begin{array}{c} 3 & y=0 \\ 2 & y=1 \\ 1 & y=2 \\ 0 & y=3 \\ \end{array}\right\}$$ And if you need some formula, it is: $$y \overset{4}{\equiv} 3x+7 \overset{4}{\equiv} -x+3 \Longrightarrow x+y \overset{4}{\equiv} 3 \Longrightarrow \left\{ \begin{array}{c} y \overset{4}{\equiv} 3-x \\ x \overset{4}{\equiv} 3-y \\ \end{array} \right\}$$