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Question Number 22896 by jazary last updated on 23/Oct/17

how can demonstred  17^(4n+1) +3×9^(2n+1) ≡0[11]

$${how}\:{can}\:{demonstred} \\ $$$$\mathrm{17}^{\mathrm{4}{n}+\mathrm{1}} +\mathrm{3}×\mathrm{9}^{\mathrm{2}{n}+\mathrm{1}} \equiv\mathrm{0}\left[\mathrm{11}\right] \\ $$

Commented by Rasheed.Sindhi last updated on 24/Oct/17

For n=1  17^5 +3×9^3 =1422044  But 11 ∤ 1422044  Hence the given congruence is  generally false

$$\mathrm{For}\:\mathrm{n}=\mathrm{1} \\ $$$$\mathrm{17}^{\mathrm{5}} +\mathrm{3}×\mathrm{9}^{\mathrm{3}} =\mathrm{1422044} \\ $$$$\mathrm{But}\:\mathrm{11}\:\nmid\:\mathrm{1422044} \\ $$$$\mathrm{Hence}\:\mathrm{the}\:\mathrm{given}\:\mathrm{congruence}\:\mathrm{is} \\ $$$$\mathrm{generally}\:\mathrm{false} \\ $$

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