Question Number 113073 by bobhans last updated on 11/Sep/20
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{n}\:\mathrm{for}\: \\ $$$$\mathrm{which}\:\mathrm{5}^{\mathrm{n}} +\mathrm{1}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7} \\ $$
Answered by john santu last updated on 11/Sep/20
$$\:\:{find}\:{all}\:{positive}\:{integers}\:{n}\:{such} \\ $$$${that}\:\mathrm{5}^{{n}} \:+\:\mathrm{1}\:{divisible}\:{by}\:\mathrm{7} \\ $$$$\left({sol}\right):\:\mathrm{5}^{{n}} +\mathrm{1}\:\equiv\:\mathrm{0}\:\left({mod}\:\mathrm{7}\:\right) \\ $$$${This}\:{can}\:{be}\:{rewritten}\:\mathrm{5}^{{n}} \equiv−\mathrm{1}\equiv\mathrm{125}\:\left({mod}\:\mathrm{7}\right) \\ $$$${However}\:{by}\:{Fermat}'{s}\:{Little}\:{Theorem} \\ $$$$\mathrm{5}^{\mathrm{6}} \:\equiv\mathrm{1}\:\left({mod}\:\mathrm{7}\right)\:.\:{Morever}\:{it}\:{easy}\:{to} \\ $$$${check}\:{that}\:\mathrm{5}^{\mathrm{2}} \neq\:\mathrm{1}\left({mod}\:\mathrm{7}\right)\:{and}\: \\ $$$$\mathrm{5}^{\mathrm{3}} \:\neq\:\mathrm{1}\:\left({mod}\:\mathrm{7}\right).\:{Hence}\:{we}\:{must}\: \\ $$$${have}\:{n}\:\equiv\:\mathrm{3}\:\left({mod}\:\mathrm{6}\right)\:{or}\:{n}=\mathrm{3}+\mathrm{6}{k}\:,\:{k}\in\mathbb{Z}^{+} \\ $$$$\:\:\:\frac{{JS}}{{a}\:{math}\:{farmer}} \\ $$
Commented by bemath last updated on 11/Sep/20
$$\checkmark\mathrm{thank}\:\mathrm{you} \\ $$