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Find all of the solutions of $4(x,y) = (1,2)$ in $\mathbb{Z}_3$ x $\mathbb{Z}_8$. What I have so far:

$4(x,y) = (1,2) \implies (4x,4y) = (1,2) \implies 4x = 1 \pmod 3$ and $4y = 2 \pmod 8$.

Bill Dubuque
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1 Answers1

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Remember that you are in $\mathbb{Z_3}$ for $x$. Thus $x \in \{0,1,2\}$ and similarly $y \in \{0,1,2,3,4,5,6,7\}$. Thus by checking case by case we have: $x = 1$. Observe that $4y = 2 \pmod 8 \implies 2y = 1\pmod 4$, and this equation has no solution in $y$. Thus there is no solution at all.