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Show that there are infinitely many primes $p$ with $$p^5 \equiv 5 \pmod 6$$

I am very confused by this question.

I am familiar with Euclid's proof of there being infinitely many primes and can see that most proofs of this nature run along a similar path.

For instance, most usually involve assuming a limited number of primes of this type and then multiplying them all together and adding a number to this and this leads to a contradiction later on.

For this problem, I have begun by also assuming there are only a finite number of primes with the above property. However, due to the index of 5, I don't see how I can simply multiply all the primes in the finite set together to figure this out. Multiplying two to more numbers with a remainder of 5 on division by six together always leads to a remainder of either 1 if the index is odd and 5 if it is even.

I do not know where I can go any further with my analysis.

Any hints at how I can tackle this question would be greatly appreciated.

VIVID
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Pied 3.14per
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    Simplify things : if $p \equiv 5 \mod 6$ then $p^5 \equiv 5 \mod 6$. So just prove that there are infinitely many primes of the form $6k-1$. To do that, first show that every number of the form $6k-1$ must have a prime factor of the form $6n-1$. Then attempt a Euclid-type proof. – Sarvesh Ravichandran Iyer Oct 28 '20 at 11:03
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    What are the primes modulo $6$ and what are their fifth powers modulo $6$? It might be easier to write $5\equiv -1$ since $-1$ is a fifth power. – Mark Bennet Oct 28 '20 at 11:03
  • By a simple generalization of Euclid's proof there are infinitely many primes $\equiv -1\pmod{6},$ (put $,S = 6\Bbb N +1,\ c(n) = 6n-1),,$ which necessarily satisfy $,p^5\equiv (-1)^5\equiv 5\pmod{6}\ \ $ – Bill Dubuque Oct 28 '20 at 12:39
  • Ah yes I see what you're getting at there thank you. It seems simple once you know how:) – Pied 3.14per Oct 28 '20 at 14:58

1 Answers1

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We know that $$p \equiv \pm1 \pmod{6}$$ for every prime $p>3$.

Also, one can verify (see below) that, there are infinitely many primes such that $$p \equiv -1 \pmod {6} \tag 1$$

Raising to the $5^\text{th}$ power, we observe that prime such that

$$p^5 \equiv -1 \equiv 5 \pmod{6}$$ are infinite, as well.

Proof of $(1)$: Assume there are only finitely many primes such that $$p \equiv -1 \pmod {6}$$ Let $P = \{p_1,p_2,\dots,p_n\}$ be all such primes. Now, we will look at the following number: $$Q = 6p_1p_2\dots p_n - 1$$ Since, it's asserted not to be prime and since $\forall p\in P: p \nmid Q$, and $2\nmid Q$, $3\nmid Q$, we have that all prime factors of $Q$ are of the form $6k + 1$. Then, $Q$ must be of the form $6k+1$, as well. However, this contradicts the fact that $$Q \equiv -1 \pmod{6}$$ by definition. $\square$

VIVID
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