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I had a question asking when is 3 a seventh power modulo a prime $p$ if $p=1(7)$. However, I tried to find just one example using mathematica but I went up to primes in the thousands and I still couldn't find an example, so I began thinking this was a trick question.

We were learning about quadratic reciprocity, but I wasn't sure how to extend what we learned about quadratic powers to seventh powers.

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    Since, in ${\mathbb{Z}}/757\mathbb Z$, we have $2^{84}=3$, we have $(2^{12})^7=3$, in other words $3 \equiv 311^7 \pmod{757}$.2012-11-03

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Based on what Jack and countinghaus said, I started looking for primes such that $3^{\frac{p-1}{7}}\equiv 1 (p)$ and found actually that $p=757$ works!

Others that work include: 1583, 1597, 2843, 2927.

However, I'm not sure if this will always work.

I know based on what they said, if $3$ is going to be a seventh root then $3^{\frac{p-1}{7}}\equiv 1 (p)$ must be true, but could someone tell me if the converse is true?

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    @Andrew: Yes, my comment was intended to describe the (sort of) "general" case, and you have used it for our particular setting.2016-05-23
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There is much classical work on this and related questions. You can find pointers to the literature from the introduction of Stanislav Jakubec's $ $ Criterion for 3 to be eleventh power, $ $ Acta Mathematica et Informatica Universitatis Ostraviensis (1995), Vol. 03, 1, excerpted below enter image description here enter image description here enter image description here enter image description here

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$3$ is not a seventh power $\pmod{29}$ since: $ 3^{\frac{29-1}{7}}\equiv 81\equiv -6\not\equiv 1 \pmod{29}.$ Another way to state the same is that the only seventh powers in $\mathbb{Z}_{/29\mathbb{Z}}^*$ are $\pm 1$ and $\pm 12$.

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    Is the converse true? I.e. if $3^{\frac{p-1}7}\equiv 1 (p)$ is it true then $x^7 \equiv 3 (p)$ always has a solution?2012-11-03