I am trying to solve the equivalence:
$2x + 4y \equiv 6 \,\, (mod \,\, 8)$
So far, I have: $4y \equiv -2x + 6 \,\, (mod \,\, 8)$
And I am trying to find the inverse of $4 \,\, mod \,\, 8$, so I can multiply that on both sides of the equation, leaving the coefficient of $y$ to be 1. However, I am struggling to find the inverse of $4 \,\, mod \,\, 8$, since I can't find a number to multiply to 4 that will make it $8x + 1$, since $4$ is even.
Here 4 is not invertible mod 8.
Generally, an integer $a$ is invertible mod $n$ if and only if $\gcd(a,n)=1$.
In $2(x+2y)\equiv 2\cdot 3\mod 8$ one cannot use the shortening rule, which would lead to $x+2y\equiv 3\mod 8$ (shortening by 2), since 2 is a zero-divisor mod 8 which means that $2\cdot 4=0 \mod 8$.
$ > (2x + 4y) \equiv 6 \pmod{8}$
$a \equiv b \pmod{c} \iff c ~| ~(a - b) ~: ~a,b,c \in \Bbb{Z}, ~c \neq 0$.
Therefore, the problem is converted to :
find all $(x,y) \in \Bbb{Z^2}$ such that $8 ~| ~(2x + 4y - 6).$
However, $8 ~| ~(2x + 4y - 6) \iff 4 ~| ~(x + 2y - 3).$
Therefore, the problem is converted to :
find all $(x,y) \in \Bbb{R^2}$ such that $(x + 2y) \equiv 3 \pmod{4}.$
For such a simple problem, there are a variety of approaches. My approach is:
$x$ will have form $4m + r$, where $r \in \{0,1,2,3\}.$
$y$ will have form $4n + s$, where $s \in \{0,1,2,3\}.$
Then, $r + 2s \equiv 3\pmod{4} \implies r \equiv 3 - 2s \pmod{4}$.
Clearly, $s \in \{0,2\} \implies r = 3$
and $s \in \{1,3\} \implies r = 1.$
Thus, the full set of solutions is :
$\{(x,y) \in \Bbb{Z^2} ~: $
$[x \equiv 1\pmod4 ~\text{and}~ y \equiv 1\pmod4]~~$ or
$[x \equiv 1\pmod4 ~\text{and}~ y \equiv 3\pmod4]~~$ or
$[x \equiv 3\pmod4 ~\text{and}~ y \equiv 0\pmod4]~~$ or
$[x \equiv 3\pmod4 ~\text{and}~ y \equiv 2\pmod4]\}.$
Note that the above specification may be shortened by recognizing that $y$ even if and only $y$ is equivalent to either $0$ or $2, \pmod{4}.$
