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Given the congruence $x^3 \equiv a$ (mod p), where $p \geq 5$ is a prime and gcd($a,p$)=1, prove the following

a. If $p \equiv 1$ (mod 6), then the congruence has either no solutions or three incongruent solutions modulo p.

b.If $p \equiv 5$ (mod 6), then the congruence has a unique solution modulo p.

How to solve it...by using the theory of indices.

文奭允
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1 Answers1

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Let $g$ be a primitive root modulo $p$. Write $x\equiv g^X$ and $a\equiv g^A\pmod p$. (That is, $x$ has index $X$ and $a$ has index $A$.) Then the congruence is equivalent to $$3X\equiv A\pmod{p-1}\ .$$

  • If $p\equiv5\pmod6$ then $\gcd(3,p-1)=1$ and so the congruence has a unique solution.
  • If $p\equiv1\pmod6$ then $\gcd(3,p-1)=3$ and so the congruence has three solutions (if $A$ is a multiple of $3$) or none (if not).
David
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