Show that $x^3 \equiv 3 \pmod{p}$ is solvable if $p$ is of the form $6m+5$.

Question

The question is:

Show that $x^3 \equiv 3 \pmod{p}$ is solvable if $p$ is of the form $6m+5$. How many solutions are there?

Any help/hints would be appreciated!

Answer 1

It as simple as that:

The map $(\mathbb Z/p\mathbb Z)^* \to (\mathbb Z/p\mathbb Z)^*, x \mapsto x^3$ is injective, since there is no element of order $3$ in a group of size $6m+4$.

An injective self-mapping on a finite set is bijective. This also answers the second question.

Of course we can easily broaden the result without any more work: Whenver $m$ and $p-1$ are co-prime, we obtain a unique solution for the equation $x^m=a \pmod p$ for any $a$.

Answer 2

A broad hint: try Fermat's little theorem:

• What is $3^p$ congruent to,$\bmod p$?
• What is $3^{p-1}$ congruent to?
• Then what's $3^{2p-1}$ congruent to?
• In terms of $m$, what is $2p-1$?

As for the number of solutions, the fundamental theorem of algebra should help...