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Let $p$ be an odd prime. Prove.

$p \geq 5$ is of the form $6k + 1$ or of the form $6k + 5$ for some nonnegative integer $k$.

The solution says that there are 3 cases: $a = 3k$, $a = 3k+ 1$, and $a = 3k + 2$, but I can't figure out how they got those cases. Thanks for your help.

Martin Sleziak
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Alexis112
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  • What are the three possible remainders when you divide a number by $3$? (Remember, $0$ is a possible value.) – Ethan Bolker Oct 16 '16 at 17:49
  • Welcome to the site! An advice: it is strongly encourage to typeset questions and answers by using MathJax (http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference),. Not only this is for clarity, but it also improve the probability of getting an answer. – Kolmin Oct 16 '16 at 17:52
  • Thanks, I will do that next time. I was confused because I wasn't sure why they were choosing to divide the set of positive integers into 3, but I see now that it was to get the answer into the form we were looking for. Thanks for the quick response! – Alexis112 Oct 16 '16 at 18:12

2 Answers2

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Youcertainly can write $p=6k+r$ with $k\in\Bbb N_0$ and $r\in\{0,1,2,3,4,5\}$ (division with remainder). What can you say about $p$ if $r=0$, or $r=2$, or $r=3$, or $r=4$?

Hagen von Eitzen
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Hagen's answer is a good one, but I thought I might point out something more general.

First pick $n$ primes. Let's say the first two. Now every prime number is coprime to both of these numbers. This means that we require $$p \not\equiv_2 0$$ $$p \not\equiv_3 0$$ and consequently, $$p \not\equiv_6 0,2,3,4$$ $$p \equiv_6 1 \text{ or } 5$$ We may extend this further to any number of primes that we choose. Let's pick $3$ primes instead now. $2 \cdot 3 \cdot 5 = 30$, so that will be the final modular base.

$$p \not\equiv_2 0$$ $$p \not\equiv_3 0$$ $$p \not\equiv_5 0$$ and consequently, $$p \not\equiv_{30}0,2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28$$ $$p \equiv_{30} \quad\text{ one of } \quad 1,7,11,13,17,19,23,29$$ The same process may be followed not just for a choice of consecutive primes, but any choice of numbers at all. The modular base can be reduced to the lowest common multiple of the choice of numbers.

The most interesting cases I suppose are choices of two small primes, since these give small modular bases with many of the possibilities removed. The simplest case is $2,3$. It's kind of an extension to choosing just one prime, $2$, and saying that $p \equiv_2 1$.

Myridium
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