Question Number 136516 by metamorfose last updated on 22/Mar/21
$${find}\:{all}\:{integers}\:\left({x},{y}\right)\::\:\mathrm{5}^{{x}} =\mathrm{3}^{{x}} +\mathrm{2021}{y} \\ $$
Answered by MJS_new last updated on 23/Mar/21
$$\mathrm{solution}:\:{x}=\mathrm{322}{n}\wedge{y}=\frac{\mathrm{5}^{\mathrm{322}{n}} −\mathrm{3}^{\mathrm{322}{n}} }{\mathrm{2021}}\forall{n}\in\mathbb{N} \\ $$
Commented by MJS_new last updated on 23/Mar/21
$$\mathrm{yes} \\ $$$$\mathrm{3}^{{x}} =\mathrm{2021}{n}_{\mathrm{3}} +{r}_{\mathrm{3}} \\ $$$$\mathrm{5}^{{x}} =\mathrm{2021}{n}_{\mathrm{5}} +{r}_{\mathrm{5}} \\ $$$${y}=\frac{\mathrm{5}^{{x}} −\mathrm{3}^{\mathrm{3}} }{\mathrm{2021}}\:\Rightarrow\:{r}_{\mathrm{3}} ={r}_{\mathrm{5}} \\ $$$$\mathrm{obviously}\:\mathrm{this}\:\mathrm{is}\:\mathrm{given}\:\mathrm{for}\:{x}=\mathrm{0}\:\Rightarrow\:{r}_{\mathrm{3}} ={r}_{\mathrm{5}} =\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{no}\:\mathrm{other}\:\mathrm{chance}\:\mathrm{but}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{next} \\ $$$${x}\:\mathrm{by}\:\mathrm{trying}... \\ $$
Commented by Rasheed.Sindhi last updated on 23/Mar/21
$$\mathcal{S}{ir},\:{did}\:{you}\:{make}\:{a}\:{program}\:{for}\:{this}? \\ $$
Commented by Rasheed.Sindhi last updated on 23/Mar/21
$$\mathcal{T}{h}\alpha{n}\Bbbk{s}\:\mathcal{S}{ir}! \\ $$