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## Homework Statement

Let [itex] p [/itex] be a prime number such that [itex]p \equiv 1 (mod 3) [/itex]

Let [itex]a[/itex] be an integer not divisible by [itex]p[/itex]. Show that if the congruence [itex] x^3 \equiv a (mod p) [/itex] has a solution then

[tex] a^{\frac{p - 1} {3}} \equiv 1 (mod p) [/tex]

## The Attempt at a Solution

Right, I'm not sure how to prove this. I've got a couple of ideas at how things might relate to one another.

I can see that [itex]gcd(a,p) = 1[/itex] and also that

[itex] \frac{p-1}{3} = k [/itex] for some integer k. I think this relates to the power of a in the equation.

[tex] a^{\frac{p - 1} {3}} \equiv 1 (mod p) [/tex].

Could you also apply fermat's little theorem in some way to this?

Also, I don't know what to do with [itex] x^3 \equiv a (mod p) [/itex].

Could someone give me a hint or two in how to prove this? It would be much appreciated.

Thanks

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