Question Number 144693 by mathdanisur last updated on 27/Jun/21
Answered by Rasheed.Sindhi last updated on 28/Jun/21
![All natural numbers fall into the following 6 catagories with respect to modulo 6 : 6k,6k+1,6k+2,6k+3,6k+4,6k+5 Obviously 6k_(=6k) ,6k+2_(=2(3k+1)) ,6k+3_(=3(2k+1)) ,6k+4_(=2(3k+2)) are composite for k>0 ∴ All Primes≥5 fall into the two catagories: 6k+1 & 6k+5 [k∈{0,1,2,...}] ^• If q=6k+1 (6k+1)^2 mod12=(36k^2 +12k+1)mod12 =12(3k^2 +1)+1mod12=1 ^• If q=6k+5 (6k+5)^2 mod12=(36k^2 +60k+25)mod12 =( 12(3k^2 +5k+2)+1 )mod12=1 ∴ For any prime q≥5 q^2 mod 12 =1](Q144709.png)
$${All}\:{natural}\:{numbers}\:{fall}\:{into}\:{the} \\ $$$${following}\:\mathrm{6}\:{catagories}\:{with}\:{respect} \\ $$$${to}\:{modulo}\:\mathrm{6}\:: \\ $$$$\mathrm{6}{k},\mathrm{6}{k}+\mathrm{1},\mathrm{6}{k}+\mathrm{2},\mathrm{6}{k}+\mathrm{3},\mathrm{6}{k}+\mathrm{4},\mathrm{6}{k}+\mathrm{5} \\ $$$${Obviously}\:\underset{=\mathrm{6}{k}} {\mathrm{6}{k}},\underset{=\mathrm{2}\left(\mathrm{3}{k}+\mathrm{1}\right)} {\mathrm{6}{k}+\mathrm{2}},\underset{=\mathrm{3}\left(\mathrm{2}{k}+\mathrm{1}\right)} {\mathrm{6}{k}+\mathrm{3}},\underset{=\mathrm{2}\left(\mathrm{3}{k}+\mathrm{2}\right)} {\mathrm{6}{k}+\mathrm{4}} \\ $$$${are}\:{composite}\:{for}\:{k}>\mathrm{0} \\ $$$$\therefore\:{All}\:{Primes}\geqslant\mathrm{5}\:{fall}\:{into}\:{the}\:{two} \\ $$$$\:{catagories}:\:\mathrm{6}{k}+\mathrm{1}\:\&\:\mathrm{6}{k}+\mathrm{5}\:\left[{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},...\right\}\right] \\ $$$$\:^{\bullet} {If}\:\:{q}=\mathrm{6}{k}+\mathrm{1} \\ $$$$\left(\mathrm{6}{k}+\mathrm{1}\right)^{\mathrm{2}} {mod}\mathrm{12}=\left(\mathrm{36}{k}^{\mathrm{2}} +\mathrm{12}{k}+\mathrm{1}\right){mod}\mathrm{12} \\ $$$$=\mathrm{12}\left(\mathrm{3}{k}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{1}{mod}\mathrm{12}=\mathrm{1} \\ $$$$\:^{\bullet} {If}\:\:{q}=\mathrm{6}{k}+\mathrm{5} \\ $$$$\left(\mathrm{6}{k}+\mathrm{5}\right)^{\mathrm{2}} {mod}\mathrm{12}=\left(\mathrm{36}{k}^{\mathrm{2}} +\mathrm{60}{k}+\mathrm{25}\right){mod}\mathrm{12} \\ $$$$=\left(\:\mathrm{12}\left(\mathrm{3}{k}^{\mathrm{2}} +\mathrm{5}{k}+\mathrm{2}\right)+\mathrm{1}\:\right){mod}\mathrm{12}=\mathrm{1} \\ $$$$\therefore\:\:{For}\:{any}\:{prime}\:{q}\geqslant\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{q}^{\mathrm{2}} \:{mod}\:\mathrm{12}\:\:=\mathrm{1} \\ $$
Commented by mathdanisur last updated on 28/Jun/21
$${perfect}\:{Sir}\:{thanks} \\ $$