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I have the following assignment: Let $p$ be a prime number, show that if $p\equiv_{3}1$, then $p\equiv_{6}1$

I am having trouble making an approach to the solution.I appreciate any help. I can´t make the modulo to change to show the equivalence. I only proved that $p^2=3L+1$, equivalent to $p^2\equiv_{3}1$.

Thank you in advance.

Ronald Fuentes
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2 Answers2

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Let $p$ be a prime with $p\equiv 1\pmod 3$. Suppose toward a contradiction that $p\not\equiv 1\pmod 6$. Since $p\equiv 1\pmod 3$ we must then have that $p\equiv 4\pmod 6$. But then $p$ is even and since $p$ is prime $p = 2$. But $2\not\equiv 1\pmod 3$, a contradiction.

nbritten
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  • Thank you so much!! – Ronald Fuentes Dec 08 '19 at 18:03
  • Why if since ≡1(mod3) we must then have that ≡4(mod6)? – Ronald Fuentes Dec 08 '19 at 18:09
  • @FrangoFuentes $p\equiv 1\pmod 3$ says $p$ is $1$ more than a multiple of $3$, so it is either $1$ or $4$ than a multiple of $6$. We're assuming $p\not\equiv 1 \pmod 6$, so it must be $4$ more than a multiple of $6$. – nbritten Dec 08 '19 at 18:11
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    @FrangoFuentes : If you want to spell it out more explicitly to ensure that you're reasoning correctly, you can try all $6$ possibilities for the remainder of $p$ modulo $6$, and see that only $1$ is possible. That is, $p=6k$ or $p=6k+1$ or … or $p=6k+5$. … – Toby Bartels Dec 08 '19 at 18:42
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    … You can rewrite each of these equations as $p=3(2k)+\text{something}$ or $p=3(2k+1)+\text{something}$ to see what $p$ is modulo $3$, and only $2$ of these will give the correct answer ($1$) to that. You can also rewrite them all as $p=2(3k)+\text{something}$ etc, and only $3$ of them will come out odd (which is required of every prime number besides $2$, as in nbritten's argument). And only $1$ will survive both challenges, which is the $p=6k+1$ possibility. – Toby Bartels Dec 08 '19 at 18:42
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Hint: $ $ it's easy as $\,a,m,n = 1,2,3\,$ below, for $\,p\,$ any integer.

CCRT $ $ For coprime $\,m,n\!:\,\ p\equiv a\pmod{\!mn}\iff \begin{align}&p\equiv a\!\!\pmod{\!m}\\ &p\equiv a\!\!\pmod{\!n}\end{align}$

since $\, mn\mid p\!-\!a\iff m,n\mid p\!-\!a\,$ by $\,m,n\,$ coprime (so lcm = product, or by Euclid's Lemma)

Remark $\ $ This is CCRT = Constant case of CRT = Chinese Remainder Theorem - a ubiquitous result well worth commiting to memory. You can find a handful of complete proofs in the linked answer if the above hint doesn't suffice.

Bill Dubuque
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