Question: How can I solve: $min\{x \in \mathbb N_0 \quad |x \cdot 714 \equiv 1292 \mod 1972 \} $ ?
I only know about:
$x \cdot a \equiv _m b \Rightarrow m|x \cdot a - b$
different way of notation:
$min\{x \in \mathbb N_0 \quad | x \cdot 714 \equiv_{1972} 1292\}$
How to go on?
I appreciate every hint.
As $714=34\cdot21$, $1292=34\cdot 38$, $1972=34\cdot 58$, this is equivalent to solving $21 x\equiv 38\mod 58$.
We have to find the inverse of $21$ modulo $58$. The tool for this is the Extended Euclidean algorithm: $$\begin{array}{rrrl} \hline r_i&u_i&v_i&q_i\\ \hline 58&0&1\\ 21&1&0&2\\ \hline 16&-2&1&1\\ 5&3&-1&3\\1&-11&4\\ \hline \end{array}$$ Thus the inverse of $21$ mod $58$ is $-11\equiv47$. The solution is $$x\equiv47\cdot 38\equiv (-11)(-20)\equiv 46\mod 58. $$
Let's say you want to solve $ax \equiv b \mod{m}$. If $gcd(a,m)=1$, then there is a unique solution. (You should be able to get this solution)
Now if $gcd(a,m) = d$, then solution exists iff $b$ divides $d$. You should try to reduce this to a equation between $\frac{a}{d}, \frac{b}{d} $ and $\frac{m}{d}$. From this solution , you will get mutiple solutions for the initial equation. Can you easily find the min of that?
Further hint for the reduction - Solving $ax \equiv b \mod{m}$ is same as solving $ax = b + my$.
