Find all integer solutions of $2n \equiv 12 \bmod 19$
So I have re-arranged to: $2x-19y=12$ and by the extended Euclidean Algorithm, I get $$x=1 \ $$ $$y=-9$$
However, this is how far I was able to get to and not sure what follow past this point? What exactly are we looking for?
Answer
I would say there are infinitely many. Another way to think of $2n \equiv 12 \pmod{19}$ is that $2n - 12 = 19k$ for some $k \in \mathbb{Z}$. So $n = \frac{19k + 12}{2}$. Thus any even $k$ will produce an integer solution.
EDIT: It might be better to think in terms of equivalence classes for the sake of negative $k$ values. So maybe $[n] = \bigl[\frac{19k + 12}{2}\bigr]$ in $\mathbb{Z}_{19}$. For example: If $k = -2$, then $[n]_{19} = [-13]_{19} = [6]_{19}$.
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