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Show there are infinitely many primes of the form $5k+1$. This question already has answers here: Infinitely many primes $5n+1$, I am trying to follow Thomas Andrew's answer.

Why does $m^5-1\equiv 0$ imply $p\equiv 1 \pmod{5}$, because $m\neq 1 \pmod{5}$.

Bill Dubuque
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Shean
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    In your linked question, the answer before explains it: "we have that the order of $a$ in $\mathbb{F}_p^*$ is five, hence $5\mid (p-1)$. So $p\equiv 1\bmod 5$. – Dietrich Burde Feb 04 '23 at 18:56
  • There is also an old, more general thread. May be won't help you much though :-( – Jyrki Lahtonen Feb 04 '23 at 19:04
  • By the way, it's not necessary to separately prove that $n$ is not prime, since if it were, then $n$ itself would be a new prime congruent to $1 \pmod 5$. All you need is that $n > 1$ and every prime dividing $n$ is $1 \pmod 5$. – Ravi Fernando Feb 04 '23 at 19:18
  • @DietrichBurde Is this because $a^{p-1}\equiv 1 \pmod{p}$ and $a^5\equiv 1 \pmod{p}$? Does this complete the proof? – Shean Feb 04 '23 at 19:28
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    You need that $a$ has order $5$. Then the order of an element divides the order of the group (which is $p-1$ here), by Lagrange. So your solution is not complete if you don't mention Lagrange :) – Dietrich Burde Feb 04 '23 at 19:32

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To answer "Why $m \neq 1 (\mod p)$ implies $p = 1 (\mod5)$":

$m \neq 1 (\mod p)$ but $m^5 = 1 (\mod p)$. Therefore, the subgroup $<m> \subset (\mathbb{Z} / p \mathbb{Z})^*$ has order 5. The order of $(\mathbb{Z} / p \mathbb{Z})^*$ must be a multiple of $5$ by Lagrange's theorem.

The order of the $(\mathbb{Z} / p \mathbb{Z})^*$ is $p-1$, so the $p = 1 (\mod 5)$.

David Lui
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