Find all the solutions of the equation $x^3=2$ in the field $\mathbb F_{25}$

Question

We have $\mathbb{Z}_5$ the ring of integers modulo 5, which through the element $\sqrt{2}$ will be extended to a field $\mathbb{K}$ with 25 elements. I have to find all solutions to the equation $x^3=2$ in $\mathbb{K}$.

As a start, I wrote up $\mathbb{K}=\{a+b\sqrt{2}\|a,b \in\mathbb{Z_5} \}$ according to information I read on Wikipedia and I tried to do a table with all this 25 elements and finding $x^3$ for each of them. I haven't found any solution until now.

My questions:

1. is the way i described $\mathbb{K}$ correct and especially why.
2. since the elements of $\mathbb{Z_5}$ are congruence classes, does this imply something for my equation, if the solution doesn't come up in my table.
3. if all of what I am doing is okay, how can I do it in a more elegant way, than simply trying all of them out?

Important: Please, I only want some subtle hints and a bit of a direction, I really want to solve it myself.

Also: I am in my first semester ever in Algebra so my understanding is still not that deep.

I thank everyone in advance for the help

Annalisa

Answer 1

You want to find all $a,b\in\Bbb{F}_5$ such that $(a+b\sqrt{2})^3=2$. Expanding shows that $$2=(a+b\sqrt{2})^3=a^3+3a^2b\sqrt{2}+6ab^2+2b^3\sqrt{2}=a(a^2+6b^2)+b(3a^2+2b^2)\sqrt{2}.$$ So you want to simultaneously solve the equations $$a(a^2+6b^2)=2\qquad\text{ and }\qquad b(3a^2+2b^2)=0,$$ for $a,b\in\Bbb{F}_5$. Can you take it from here?

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Note that $\Bbb{F}_5^{\times}$ is a cyclic group of order $4$, so every element is a perfect cube. In particular $x^3=2$ has a solution in $\Bbb{F}_5$. Similarly $\Bbb{F}_{25}^{\times}$ is a cyclic group of order $24$, so every element either has $3$ cube roots, or no cube roots in $\Bbb{F}_{25}^{\times}$. It follows that $x^3=2$ has precisely $3$ solutions in $\Bbb{F}_{25}$.

Answer 2

Let $x = a + b\sqrt 2$ where $a,b \in \mathbb Z_5$

So $x^3 = (a + b\sqrt 2)^3 = a^3 + 3a^2b\sqrt 2 + 6ab^2 + 2b^3\sqrt 2 =(a^3+6ab^2) + (3a^2b + 2b^3)\sqrt 2 =2$

So $a^3 + 6ab^2 = a^3 + ab^2 = 2$ and $3a^2b + 2b^3=2b^3 - 2a^2b = 0$

The second equation we have $2b(b^2 - a^2) = 2b(b-a)(b+a) = 0$.

So we could have:

$b = 0$ and $a^3 =2$ but and $1,2,3,4$ cubed are $1,3,2, 4$ so $x= 3 + 0\sqrt 2 = 2$ will work (as $3^3=27 =2$).

$a=b$ and $2a^3=2$ and $a^3=1$ so $a=1$ and $x = 1 +\sqrt 2$ (and indeed $(1+\sqrt 2)^3 = 1 + 3\sqrt 2 + 6+ 2\sqrt 3 = 7 + 5\sqrt 2= 2 + 0\sqrt 2 = 2$.)

$a=-b$ so $a^3 + a^3 = 2$ and is the same... $a=1$ and $b=-1$ so $x=1-\sqrt 2$ (and indeed $(1-\sqrt 2)^3 =1 - 3\sqrt 2 + 6-2\sqrt 3=7-5\sqrt 2=2$)