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Question Number 4942 by prakash jain last updated on 24/Mar/16

If p is a prime number greater than 5, the prove  that p mod 6 ≡1 or p mod 6≡5.  i.e. All prime numbers greater than 5 leave a  remainder of 1 or 5 when divided by 6.

$$\mathrm{If}\:{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{5},\:\mathrm{the}\:\mathrm{prove} \\ $$$$\mathrm{that}\:{p}\:\mathrm{mod}\:\mathrm{6}\:\equiv\mathrm{1}\:\mathrm{or}\:{p}\:\mathrm{mod}\:\mathrm{6}\equiv\mathrm{5}. \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{All}\:\mathrm{prime}\:\mathrm{numbers}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{5}\:\mathrm{leave}\:\mathrm{a} \\ $$$$\mathrm{remainder}\:\mathrm{of}\:\mathrm{1}\:\mathrm{or}\:\mathrm{5}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{6}. \\ $$

Commented by Yozzii last updated on 25/Mar/16

  Define the function f: P→N such that  f(p)=p (mod 6) where 0≤f(p)≤6 and p>5.  Since p∈P, ∄n∈N such that p=6n+2=2(3n+1) (>5),  p=6n+3=3(2n+1) (>5),p=6n+4=2(3n+2) (>5) or p=6n=2×3n (>5). Therefore,  we know that f(p)≠0,2,3,4 and so,  possibly f(p)=1 or f(p)=5 since for   p=6n+1 or p=6n+5, no integer factors  arise from these formulae of p.  If n is even, f(6n+1)=1 or f(6n+5)=5.  If n is odd⇒p=6(2k+1)+1=12n+7  ⇒f(12k+7)=1 or if p=6(2k+1)+5=12k+11  ⇒f(12k+11)=5.

$$ \\ $$$${Define}\:{the}\:{function}\:{f}:\:\mathbb{P}\rightarrow\mathbb{N}\:{such}\:{that} \\ $$$${f}\left({p}\right)={p}\:\left({mod}\:\mathrm{6}\right)\:{where}\:\mathrm{0}\leqslant{f}\left({p}\right)\leqslant\mathrm{6}\:{and}\:{p}>\mathrm{5}. \\ $$$${Since}\:{p}\in\mathbb{P},\:\nexists{n}\in\mathbb{N}\:{such}\:{that}\:{p}=\mathrm{6}{n}+\mathrm{2}=\mathrm{2}\left(\mathrm{3}{n}+\mathrm{1}\right)\:\left(>\mathrm{5}\right), \\ $$$${p}=\mathrm{6}{n}+\mathrm{3}=\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)\:\left(>\mathrm{5}\right),{p}=\mathrm{6}{n}+\mathrm{4}=\mathrm{2}\left(\mathrm{3}{n}+\mathrm{2}\right)\:\left(>\mathrm{5}\right)\:{or}\:{p}=\mathrm{6}{n}=\mathrm{2}×\mathrm{3}{n}\:\left(>\mathrm{5}\right).\:{Therefore}, \\ $$$${we}\:{know}\:{that}\:{f}\left({p}\right)\neq\mathrm{0},\mathrm{2},\mathrm{3},\mathrm{4}\:{and}\:{so}, \\ $$$${possibly}\:{f}\left({p}\right)=\mathrm{1}\:{or}\:{f}\left({p}\right)=\mathrm{5}\:{since}\:{for}\: \\ $$$${p}=\mathrm{6}{n}+\mathrm{1}\:{or}\:{p}=\mathrm{6}{n}+\mathrm{5},\:{no}\:{integer}\:{factors} \\ $$$${arise}\:{from}\:{these}\:{formulae}\:{of}\:{p}. \\ $$$${If}\:{n}\:{is}\:{even},\:{f}\left(\mathrm{6}{n}+\mathrm{1}\right)=\mathrm{1}\:{or}\:{f}\left(\mathrm{6}{n}+\mathrm{5}\right)=\mathrm{5}. \\ $$$${If}\:{n}\:{is}\:{odd}\Rightarrow{p}=\mathrm{6}\left(\mathrm{2}{k}+\mathrm{1}\right)+\mathrm{1}=\mathrm{12}{n}+\mathrm{7} \\ $$$$\Rightarrow{f}\left(\mathrm{12}{k}+\mathrm{7}\right)=\mathrm{1}\:{or}\:{if}\:{p}=\mathrm{6}\left(\mathrm{2}{k}+\mathrm{1}\right)+\mathrm{5}=\mathrm{12}{k}+\mathrm{11} \\ $$$$\Rightarrow{f}\left(\mathrm{12}{k}+\mathrm{11}\right)=\mathrm{5}. \\ $$

Commented by 3 last updated on 15/May/16

h

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