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Question Number 115595 by floor(10²Eta[1]) last updated on 26/Sep/20

$$\mathrm{Prove}\mathrm{that},\mathrm{for}\mathrm{all}\mathrm{primes}\mathrm{p}>3,$$ $$13\mid {10}^{2\mathrm{p}}-{10}^{\mathrm{p}}+1$$

Answered by MJS_new last updated on 27/Sep/20

$${10}^{6n}\equiv 1\mathrm{mod}13$$ $${10}^{6n+1}\equiv 10\mathrm{mod}13$$ $${10}^{6n+2}\equiv 9\mathrm{mod}13$$ $${10}^{6n+3}\equiv 12\mathrm{mod}13$$ $${10}^{6n+4}\equiv 3\mathrm{mod}13$$ $${10}^{6n+5}\equiv 4\mathrm{mod}13$$ $$\Rightarrow$$ $$a\mathrm{mod}13-b\mathrm{mod}13+1\mathrm{mod}13=0\mathrm{mod}13$$ $$\mathrm{has}\mathrm{the}\mathrm{solutions}$$ $$a=9\wedge b=10\mathrm{or}a=3\wedge b=4$$ $$\Rightarrow a={10}^{6m+2}\wedge b={10}^{6n+1}\mathrm{or}a={10}^{6m+4}\wedge b={10}^{6n+5}$$ $$\Rightarrow$$ $$2p=6m+2\wedge p=6n+1\mathrm{or}2p=6m+4\wedge p=6n+5$$ $$\left(1\right)$$ $$p=6n+1\Rightarrow 2p=12n+2=6\left(2n\right)+1=6m+1$$ $$\left(2\right)$$ $$p=6n+5\Rightarrow 2p=12n+10=6(2n+1)+4=6m+4$$ $$\Rightarrow \mathrm{both}\mathrm{are}\mathrm{true}$$ $$\Rightarrow$$ $$13\mid {10}^{2p}-{10}^p+1\mathrm{with}p=6n+1\vee p=6n+5$$ $$\iff p=\mathrm{any}\mathrm{uneven}\mathrm{number}\mathrm{not}\mathrm{divisible}\mathrm{by}3$$ $$\Rightarrow \mathrm{all}\mathrm{primes}\mathrm{except}2,3\mathrm{are}\mathrm{included}\mathrm{in}\mathrm{this}$$ $$\mathrm{solution}$$

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