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Question Number 115595 by floor(10²Eta[1]) last updated on 26/Sep/20
Prove that, for all primes p>3, 13∣10^(2p) −10^p +1
$$\mathrm{Prove}\:\mathrm{that},\:\mathrm{for}\:\mathrm{all}\:\mathrm{primes}\:\mathrm{p}>\mathrm{3}, \\ $$$$\mathrm{13}\mid\mathrm{10}^{\mathrm{2p}} −\mathrm{10}^{\mathrm{p}} +\mathrm{1} \\ $$
Answered by MJS_new last updated on 27/Sep/20
10^(6n) ≡1mod13 10^(6n+1) ≡10mod13 10^(6n+2) ≡9mod13 10^(6n+3) ≡12mod13 10^(6n+4) ≡3mod13 10^(6n+5) ≡4mod13 ⇒ amod13−bmod13+1mod13=0mod13 has the solutions a=9∧b=10 or a=3∧b=4 ⇒ a=10^(6m+2) ∧b=10^(6n+1) or a=10^(6m+4) ∧b=10^(6n+5) ⇒ 2p=6m+2∧p=6n+1 or 2p=6m+4∧p=6n+5 (1) p=6n+1 ⇒ 2p=12n+2=6(2n)+1=6m+1 (2) p=6n+5 ⇒ 2p=12n+10=6(2n+1)+4=6m+4 ⇒ both are true ⇒ 13∣10^(2p) −10^p +1 with p=6n+1∨p=6n+5 ⇔ p = any uneven number not divisible by 3 ⇒ all primes except 2, 3 are included in this solution
$$\mathrm{10}^{\mathrm{6}{n}} \equiv\mathrm{1mod13} \\ $$$$\mathrm{10}^{\mathrm{6}{n}+\mathrm{1}} \equiv\mathrm{10mod13} \\ $$$$\mathrm{10}^{\mathrm{6}{n}+\mathrm{2}} \equiv\mathrm{9mod13} \\ $$$$\mathrm{10}^{\mathrm{6}{n}+\mathrm{3}} \equiv\mathrm{12mod13} \\ $$$$\mathrm{10}^{\mathrm{6}{n}+\mathrm{4}} \equiv\mathrm{3mod13} \\ $$$$\mathrm{10}^{\mathrm{6}{n}+\mathrm{5}} \equiv\mathrm{4mod13} \\ $$$$\Rightarrow \\ $$$${a}\mathrm{mod13}−{b}\mathrm{mod13}+\mathrm{1mod13}=\mathrm{0mod13} \\ $$$$\mathrm{has}\:\mathrm{the}\:\mathrm{solutions} \\ $$$${a}=\mathrm{9}\wedge{b}=\mathrm{10}\:\mathrm{or}\:{a}=\mathrm{3}\wedge{b}=\mathrm{4} \\ $$$$\Rightarrow\:{a}=\mathrm{10}^{\mathrm{6}{m}+\mathrm{2}} \wedge{b}=\mathrm{10}^{\mathrm{6}{n}+\mathrm{1}} \:\mathrm{or}\:{a}=\mathrm{10}^{\mathrm{6}{m}+\mathrm{4}} \wedge{b}=\mathrm{10}^{\mathrm{6}{n}+\mathrm{5}} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{p}=\mathrm{6}{m}+\mathrm{2}\wedge{p}=\mathrm{6}{n}+\mathrm{1}\:\mathrm{or}\:\mathrm{2}{p}=\mathrm{6}{m}+\mathrm{4}\wedge{p}=\mathrm{6}{n}+\mathrm{5} \\ $$$$\left(\mathrm{1}\right) \\ $$$${p}=\mathrm{6}{n}+\mathrm{1}\:\Rightarrow\:\mathrm{2}{p}=\mathrm{12}{n}+\mathrm{2}=\mathrm{6}\left(\mathrm{2}{n}\right)+\mathrm{1}=\mathrm{6}{m}+\mathrm{1} \\ $$$$\left(\mathrm{2}\right) \\ $$$${p}=\mathrm{6}{n}+\mathrm{5}\:\Rightarrow\:\mathrm{2}{p}=\mathrm{12}{n}+\mathrm{10}=\mathrm{6}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{4}=\mathrm{6}{m}+\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{both}\:\mathrm{are}\:\mathrm{true} \\ $$$$\Rightarrow \\ $$$$\mathrm{13}\mid\mathrm{10}^{\mathrm{2}{p}} −\mathrm{10}^{{p}} +\mathrm{1}\:\mathrm{with}\:{p}=\mathrm{6}{n}+\mathrm{1}\vee{p}=\mathrm{6}{n}+\mathrm{5} \\ $$$$\Leftrightarrow\:{p}\:=\:\mathrm{any}\:\mathrm{uneven}\:\mathrm{number}\:\mathrm{not}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{all}\:\mathrm{primes}\:\mathrm{except}\:\mathrm{2},\:\mathrm{3}\:\mathrm{are}\:\mathrm{included}\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{solution} \\ $$

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