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Question Number 170046 by Rasheed.Sindhi last updated on 14/May/22

$$\mathrm{PROVE}\mathrm{THAT}$$$$6\mid (8^x-2^x)$$

Answered by JDamian last updated on 14/May/22

$$(8^x-2^x)\mathrm{mod}6=$$$$=8^x\mathrm{mod}6-2^x\mathrm{mod}6=$$$$={\left(8\mathrm{mod}6\right)}^x-2^x\mathrm{mod}6=$$$$=2^x\mathrm{mod}6-2^x\mathrm{mod}6=$$$$=0$$

Commented by Rasheed.Sindhi last updated on 15/May/22

$$\mathcal{T}han\mathcal{X}sir!$$

Answered by mr W last updated on 15/May/22

$$8^n={(2+6)}^n=\underset{k=0}{\sum ^n}{C}_k^n{6}^{n-k}{2}^k=\underset{k=0}{\sum ^{n-1}}{C}_k^n{6}^{n-k}{2}^k+2^n$$$$8^n-2^n=\underset{k=0}{\sum ^{n-1}}{C}_k^n{6}^{n-k}{2}^k$$$$\Rightarrow 6\mid 8^n-2^n$$

Commented by Rasheed.Sindhi last updated on 15/May/22

$$\mathcal{T}han\mathcal{X}sir!$$

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