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I want to prove divisibility using factoring, So i need to show that $11^{10}-1$ can be written as prime factors of 100.

This is what I've tried: $$11^{10}-1 $$ $$ (11^{5})^{2}-1$$ $$ (11^{5}-1)(11^{5}+1 )$$ $$10(11^{4}+11^{3}+11^{2}+11^1+1)(12)(11^4-11^3+11^2-11^1+1)$$ $$5*2^3*3[(11^4+11^2+1)+(11^3+11)]*[(11^4+11^2+1)-(11^3+11)]$$

I wanna write the $[(11^4+11^2+1)+(11^3+11)]*[(11^4+11^2+1)-(11^3+11)]$ as $5k$ with $k$ $\in\mathbb{Z}$.

I'd love any help/hint that would move me forward.

Dota
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  • But you already solved it in the the 5th line. You have $5$ times $2$ times (something). (In fact, you solved it in the 4th line). – Yanko Apr 29 '18 at 10:41
  • Wow, this question is closed as duplicate in just $121$ seconds. – user061703 Apr 29 '18 at 10:42
  • @Yanko Sorry I meant $100|11^{10} - 1$ not $10|11^{10} - 1$ – Dota May 02 '18 at 14:44
  • I think you should ask a new question because now (after it's been marked as a duplicated answer) we are not able to answer it. – Yanko May 02 '18 at 17:51

1 Answers1

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Hint: $$11\equiv 1 \mod 10$$ so $$11^{10}\equiv 1^{10}\equiv 1 \mod 10$$

Dr. Sonnhard Graubner
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