Question Number 4203 by Yozzii last updated on 01/Jan/16 $${I}\:{have}\:{no}\:{formal}\:{background}\:{in}\: \\ $$$${number}\:{theory},\:{but}\:{I}'{m}\:{curious} \\ $$$${of}\:{how}\:{to}\:{find}\:{positive}\:{integer}\:{solutions}\: \\ $$$$\left({x},{y},{z}\right)\:{to}\:{the}\:{equation}\:{x}^{{n}} +{y}^{{n}} ={z}^{{n}} \:{for}\: \\ $$$${n}\in\mathbb{Z}^{−} .\:{Fermat}'{s}\:{last}\:{theorem}\:{led} \\ $$$${me}\:{to}\:{this}.\:{Tell}\:{me}\:{about}\:{the}\:{cases} \\…
Question Number 69574 by Ajao yinka last updated on 25/Sep/19 Answered by MJS last updated on 25/Sep/19 $$\mathrm{36}_{\mathrm{7}} =\mathrm{27}_{\mathrm{10}} \\ $$$$\mathrm{27}_{\mathrm{10}} !=\mathrm{10}\:\mathrm{888}\:\mathrm{869}\:\mathrm{450}\:\mathrm{418}\:\mathrm{352}\:\mathrm{160}\:\mathrm{768}\:\mathrm{000}\:\mathrm{000}_{\mathrm{10}} = \\ $$$$\mathrm{36}_{\mathrm{7}}…
Question Number 69571 by Ajao yinka last updated on 25/Sep/19 Answered by MJS last updated on 25/Sep/19 $${m}^{\mathrm{6}} +\mathrm{375}{n}^{\mathrm{2}} {m}^{\mathrm{2}} +{n}^{\mathrm{6}} −\mathrm{1953125}=\mathrm{0} \\ $$$${m}=\sqrt{{t}} \\…
Question Number 69569 by Ajao yinka last updated on 25/Sep/19 Answered by naka3546 last updated on 25/Sep/19 Commented by Rasheed.Sindhi last updated on 26/Sep/19 $${I}\:{think}\:'{positive}\:{integer}'\:{applies}\:{to}…
Question Number 69567 by Ajao yinka last updated on 25/Sep/19 Commented by Prithwish sen last updated on 25/Sep/19 $$\mathrm{Let}\:\mathrm{us}\:\mathrm{assume}\:\mathrm{gcd}\left(\mathrm{m},\mathrm{n}\right)=\mathrm{1} \\ $$$$\mathrm{then}\:\emptyset\left(\mathrm{mn}\right)=\emptyset\left(\mathrm{m}\right)\emptyset\left(\mathrm{n}\right) \\ $$$$\therefore\:\frac{\emptyset\left(\mathrm{mn}\right)}{\mathrm{gcd}\left(\left(\emptyset\left(\mathrm{m}\right),\emptyset\left(\mathrm{n}\right)\right)\right.}\:=\:\mathrm{10} \\ $$$$\Rightarrow\frac{\emptyset\left(\mathrm{m}\right)\emptyset\left(\mathrm{n}\right)}{\mathrm{gcd}\left(\left(\emptyset\left(\mathrm{m}\right),\emptyset\left(\mathrm{n}\right)\right)\right.}\:=\:\mathrm{10}…
Question Number 134960 by bobhans last updated on 09/Mar/21 $$\mathrm{Number}\:\mathrm{theory} \\ $$A palindrome is a number that reads the same backwards as forwards, as 3141413.…
Question Number 3685 by Filup last updated on 19/Dec/15 $$\mathrm{let}\:{p}_{{i}} \:\mathrm{be}\:\mathrm{the}\:{i}^{\mathrm{th}} \:\mathrm{prime} \\ $$$$ \\ $$$$\mathrm{Does}\:\mathrm{the}\:\mathrm{fillowing}\:\mathrm{sum}\:\mathrm{converge}: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{p}_{{i}} }{{p}_{{i}+\mathrm{1}} } \\ $$$$\left({p}_{\mathrm{1}} =\mathrm{2},\:{p}_{\mathrm{2}}…
Question Number 3659 by Filup last updated on 18/Dec/15 $${p}_{{i}} \:=\:{i}\mathrm{th}\:\mathrm{prime} \\ $$$$\mathrm{let}:\:\:\:\:\:\:\rho={p}_{{n}} −{p}_{{n}−\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{is}: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\rho=\infty? \\ $$ Commented by…
Question Number 3644 by prakash jain last updated on 17/Dec/15 $$\mathrm{An}\:\mathrm{hexagon}\:\mathrm{of}\:\mathrm{unit}\:\mathrm{side}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{on}\:\mathrm{plane}. \\ $$$$\mathrm{Draw}\:\mathrm{a}\:\mathrm{square}\:\mathrm{having}\:\mathrm{the}\:\mathrm{same}\:\mathrm{area}\:\mathrm{as}\:\mathrm{the} \\ $$$$\mathrm{hexagon}\:\mathrm{using}\:\mathrm{only}\:\mathrm{unmarked}\:\mathrm{ruler}\:\mathrm{and}\: \\ $$$$\mathrm{compass}. \\ $$$$\mathrm{What}\:\mathrm{if}\:\mathrm{an}\:{n}−\mathrm{gon}\:\mathrm{with}\:\mathrm{unit}\:\mathrm{edges}\:\mathrm{is}\:\mathrm{given}? \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{always}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{square} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{area}\:\mathrm{as}\:{n}−\mathrm{gon}\:\mathrm{using}\:\mathrm{ruler} \\ $$$$\mathrm{and}\:\mathrm{compass}.…
Question Number 3630 by prakash jain last updated on 16/Dec/15 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{prime}\:\mathrm{numbers}\:{p} \\ $$$$\mathrm{such}\:\mathrm{that}\:{n}\leqslant{p}\leqslant\mathrm{2}^{{k}} {n}\:\mathrm{is}\:\geqslant\mathrm{2}^{{k}} {n}. \\ $$$$\underset{{n}\leqslant{p}\leqslant\mathrm{2}^{{k}} {n}} {\sum}{p}\:\:\geqslant\:\mathrm{2}^{{k}} {n}\:\:\:\:\:\left({p}−{prime}\right) \\ $$ Terms of Service…