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Category: Arithmetic

find-all-integers-n-for-which-13-4-n-2-1-

Question Number 94097 by Rio Michael last updated on 16/May/20 $$\mathrm{find}\:\mathrm{all}\:\mathrm{integers}\:{n}\:\:\mathrm{for}\:\mathrm{which}\:\:\mathrm{13}\:\mid\mathrm{4}\left({n}^{\mathrm{2}} +\mathrm{1}\right). \\ $$ Commented by Rasheed.Sindhi last updated on 17/May/20 $${This}\:{s}\:{equivalent}\:{to}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{13}\:\mid\left({n}^{\mathrm{2}} +\mathrm{1}\right)…

Question-28449

Question Number 28449 by A1B1C1D1 last updated on 25/Jan/18 Answered by $@ty@m last updated on 26/Jan/18 $${there}\:{are}\:{infinitely}\:{many}\:{solutions}: \\ $$$$\left(\mathrm{0},\pm\mathrm{1}\right) \\ $$$$\left(\pm\mathrm{1},\mathrm{0}\right) \\ $$$$\left(\pm\frac{\mathrm{3}}{\mathrm{5}},\pm\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\left(\pm\frac{\mathrm{5}}{\mathrm{13}},\pm\frac{\mathrm{12}}{\mathrm{13}}\right)…

n-1-1-n-n-1-

Question Number 93859 by  M±th+et+s last updated on 15/May/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$ Answered by Ar Brandon last updated on 15/May/20 $$\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\Rightarrow\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}=\underset{\mathrm{n}=\mathrm{1}}…

Question-159189

Question Number 159189 by mathlove last updated on 14/Nov/21 Answered by Rasheed.Sindhi last updated on 16/Nov/21 $${Let}\:\frac{{A}}{{B}}={x} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1};\:{x}^{\mathrm{2019}} +{x}^{−\mathrm{2019}} =? \\ $$$$\blacktriangleright{x}+\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}}…