Prove-that-for-all-primes-p-gt-3-13-10-2p-10-p-1-

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

Prove that, for all primes p > 3,

13 ∣ 10^{2 p} - 10^{p} + 1

Answered by MJS_new last updated on 27/Sep/20

10^{6 n} \equiv 1 \mod 13

10^{6 n + 1} \equiv 10 \mod 13

10^{6 n + 2} \equiv 9 \mod 13

10^{6 n + 3} \equiv 12 \mod 13

10^{6 n + 4} \equiv 3 \mod 13

10^{6 n + 5} \equiv 4 \mod 13

\Rightarrow

a \mod 13 - b \mod 13 + 1 \mod 13 = 0 \mod 13

has the solutions

a = 9 \land b = 10 or a = 3 \land b = 4

\Rightarrow a = 10^{6 m + 2} \land b = 10^{6 n + 1} or a = 10^{6 m + 4} \land b = 10^{6 n + 5}

\Rightarrow

2 p = 6 m + 2 \land p = 6 n + 1 or 2 p = 6 m + 4 \land p = 6 n + 5

(1)

p = 6 n + 1 \Rightarrow 2 p = 12 n + 2 = 6 (2 n) + 1 = 6 m + 1

(2)

p = 6 n + 5 \Rightarrow 2 p = 12 n + 10 = 6 (2 n + 1) + 4 = 6 m + 4

\Rightarrow both are true

\Rightarrow

13 ∣ 10^{2 p} - 10^{p} + 1 with p = 6 n + 1 \lor p = 6 n + 5

\Leftrightarrow p = any uneven number not divisible by 3

\Rightarrow all primes except 2, 3 are included in this

solution

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