Question Number 80643 by mr W last updated on 05/Feb/20
$${Prove}\:{that}\:{there}\:{is}\:{no}\:{solution}\:{for} \\ $$$$\mathrm{7}^{{n}} \equiv\mathrm{1}\:{mod}\left(\mathrm{35}\right)\:{with}\:{n}\in\mathbb{N}. \\ $$
Answered by MJS last updated on 05/Feb/20
$$\forall{k}\in\mathbb{N}:\:\mathrm{7}^{\mathrm{20}{k}} \equiv\mathrm{1mod}\:\mathrm{55} \\ $$
Commented by mr W last updated on 05/Feb/20
$${thanks}\:{sir}!\:{i}\:{should}\:{have}\:{asked} \\ $$$$\mathrm{7}^{{n}} \equiv\mathrm{1}\:{mod}\:\left(\mathrm{35}\right).\:{is}\:{there}\:{a}\:{solution}? \\ $$
Commented by MJS last updated on 05/Feb/20
$$\mathrm{no}\:\mathrm{solution}\:\mathrm{except}\:{n}=\mathrm{0} \\ $$$${k}\in\mathbb{N}^{\bigstar} :\:\mathrm{7}^{\mathrm{4}{k}} \equiv\mathrm{21mod}\:\mathrm{35} \\ $$$${k}\in\mathbb{N}: \\ $$$$\:\:\:\:\:\mathrm{7}^{\mathrm{4}{k}+\mathrm{1}} \equiv\mathrm{7mod}\:\mathrm{35} \\ $$$$\:\:\:\:\:\mathrm{7}^{\mathrm{4}{k}+\mathrm{2}} \equiv\mathrm{14mod}\:\mathrm{35} \\ $$$$\:\:\:\:\:\mathrm{7}^{\mathrm{4}{k}+\mathrm{3}} \equiv\mathrm{28mod}\:\mathrm{35} \\ $$
Commented by mr W last updated on 05/Feb/20
$${thank}\:{you}\:{sir}\:{for}\:{confirming}! \\ $$
Answered by mind is power last updated on 05/Feb/20
$${only}\:{n}=\mathrm{0} \\ $$$${if}\:{n}\geqslant\mathrm{1}\Rightarrow\mathrm{7}\mid\mathrm{7}^{{n}} ,\mathrm{7}\mid\mathrm{35} \\ $$$${if}\:\exists{n}\:{such}\:\mathrm{7}^{{n}} =\mathrm{1}\left(\mathrm{35}\right)\Rightarrow \\ $$$$\mathrm{7}^{{n}} =\mathrm{1}+\mathrm{35}{k}\Rightarrow \\ $$$$\mathrm{1}=\mathrm{7}\left(\mathrm{7}^{{n}−\mathrm{1}} −\mathrm{5}{k}\right)\Rightarrow\mathrm{7}\mid\mathrm{1}\:\:{absurd} \\ $$
Commented by mr W last updated on 05/Feb/20
$${thanks}\:{alot}\:{sir}!\:{that}'{s}\:{it}! \\ $$
Commented by mind is power last updated on 05/Feb/20
$${y}'{re}\:{Welcom} \\ $$