Solving $2x^{3}=3$ over $\mathbb{Z}_5$.

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Really new to dealing with $\mathbb{Z}_n$.

I'm trying to calculate $2x^{3}=3$ over $\mathbb{Z}_5$.

I did a similar question - solving $3x^{2}+x+1=0$ over $\mathbb{Z}_5$:

$$3x^{2}+x+1=0 \\\Rightarrow6x^{2}+2x+2=0\\ \Rightarrow x^{2}+2x+1=-1=4\\ \Rightarrow(x+1)^{2}=2^{2}\\ \Rightarrow x+1=\pm2=2,3$$

But I have no idea how to solve $2x^{3}=3$. It feels like it should be easier.

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There are only 5 elements to check and $0$ and $1$ are easily ruled out. So just plug in $x = 2, 3, 4$ and see which of those work.

While we're here, let me point out that in $\mathbb{Z}_n$, there is a difference between a polynomial function and a polynomial that is not seen in $\mathbb{Q}$ or $\mathbb{R}$ or $\mathbb{C}$. For example, in $\mathbb{Z}_2$, $x = x^2$ as functions because $0 = 0^2$ and $1 = 1^2$. However, as polynomials they are different because they are made up of different terms. (This is something that was confusing to me when I first learned about polynomials in $\mathbb{Z}_n$.)