# Basic field $\Bbb Z_7$ equation

trying to figure if I understand the basics of $$\Bbb Z_7$$ field $${}=\{1,2,3,4,5,6\}$$ on the following equation:

$$6x^3$$ = $$5$$

I tried to reduce the coefficients in the equation -

$$6x^3$$ = $$5$$ => $$6*6x^3 = 6*5$$ => $$36x^3 = 30$$ => mod7 => $$x^3 = 2$$

From this stage $$x^3 = 2$$. I don't know how to continue.

Thanks in advance.

• For these kinds of questions, its a good idea to write down all the squares - or in this case, cubes - modulo the prime you are working with - in this case 7. – bounceback Oct 11 '18 at 17:17
• Thanks for the good advice! I will definitely use it in the future. – OO1 Oct 11 '18 at 19:10

## 1 Answer

Hint:

Clearly $$0$$ is not a solution. On the other hand, for any $$x\ne0$$, lil' Fermat says that $$x^6=1$$, so $$(x^3)^2=1$$. What can you deduce for $$x^3$$, knowing we're in a field?

• In that case, if I try to set any number between $0-6$ in $x$ , none of them will be equals to $2$. eventually, there is a conflict in that equation. – OO1 Oct 11 '18 at 19:13
• Yes. You necessarily obtain $1$ or $-1$, since these are two obvious solutions of $y^2=1$, and in a field, a quadratic equation cannot have more than two solutions. – Bernard Oct 11 '18 at 19:24