Solve the equation $144x\equiv24\mod228$ giving all the integer and not congruent solutions.
So, there's solution for this equation if and only if $(228:144)=d\implies d|24$
By the Extended Euclidean Algorithm: $d=12 \implies 12|24$, then it can be solved.
Solving (...)
$228 = 1 \times 144 + 84$
$144 = 1 \times 84 + 60$
$84 = 1 \times 60 + 24$
$60 = 2 \times 24 + 12$
$24 = 2 \times 12 + 0$
Clearing the rests:
$84 = 228-1\times 144$
$60 = 144 \times84$
$24=84-1 \times60$
$12=60-2\times 24$
let $r,s\in\Bbb Z$ then $12 = r\times144+s\times228$
Replacing: $$12=60-2\times24$$ $$12=60-2\times[84-1\times60]$$ $$12=3\times60-2\times84$$ $$12=3\times[144-1\times84]-2\times84$$ $$12=3\times144-5\times84$$ $$12=3\times144-5[228-1\times144]$$ $$=8\times144-5\times228$$
But $24=2\times12+0 \implies 24=2\times[8\times144-5\times228]+0\implies24=16\times144-10\times228$
then, $x_0=16$ and $t={m\over{d}}={228\over{12}}=19$
Not congruent solutions: $x_0+q\times t=16+19\times q$ with $q=1,...,11$
Integer solutiuons: $m\times k + (x_0+q\times k) = 228\times k + [16 + q \times 19]$ with $k \in \Bbb Z$
I'm not sure about the solutions, so my question is: this is well done?. If not, someone can recommend some book or information to solve this?
Thanks!
$(1)$ Divide through by 12 and reduce the problem to,
$(2)$ $12x \equiv 2\mod 19$
$(3)$ Here, $\gcd(12,19) = 1$. And $1|2$ so we are good.
$(4)$ Now, $12(8)+19(-5)=1$ (by $\textbf{ext euclidean algorithm}$),
$(5)$ $\Rightarrow 1 \equiv 12(8)\mod 19 \Rightarrow x \equiv 12x (8)\mod 19$ (multiply by $\textbf{x}$).
Hence, $x\equiv 16 \mod 19$.