Question Number 173162 by dragan91 last updated on 07/Jul/22
Answered by aleks041103 last updated on 07/Jul/22
$${x}+{y}>\mathrm{0} \\ $$$$\Rightarrow{x}^{{z}} >{y}^{{z}} \\ $$$${z}\in\mathbb{N}\Rightarrow{x}>{y} \\ $$$${let}\:{x}={y}+{t},\:{t}\in\mathbb{N} \\ $$$$\left({y}+{t}\right)^{{z}} −{y}^{{z}} =\mathrm{2}{y}+{t} \\ $$$${z}=\mathrm{1}\Rightarrow\left({y}+{t}\right)−{y}={t}=\mathrm{2}{y}+{t}\Rightarrow{y}=\mathrm{0}\notin\mathbb{N} \\ $$$$\Rightarrow{z}>\mathrm{1} \\ $$$$\left({y}+{t}\right)^{{z}} −{y}^{{z}} ={y}^{{z}} \left(\left(\mathrm{1}+\frac{{t}}{{y}}\right)^{{z}} −\mathrm{1}^{{z}} \right)= \\ $$$$={y}^{{z}} \frac{{t}}{{y}}\left(\mathrm{1}+\left(\mathrm{1}+\frac{{t}}{{y}}\right)+\left(\mathrm{1}+\frac{{t}}{{y}}\right)^{\mathrm{2}} +…+\left(\mathrm{1}+\frac{{t}}{{y}}\right)^{{z}−\mathrm{1}} \right)= \\ $$$$={ty}^{{z}−\mathrm{1}} \left(\mathrm{1}+\left(\mathrm{1}+\frac{{t}}{{y}}\right)+\left(\mathrm{1}+\frac{{t}}{{y}}\right)^{\mathrm{2}} +…+\left(\mathrm{1}+\frac{{t}}{{y}}\right)^{{z}−\mathrm{1}} \right) \\ $$$$\Rightarrow\left({y}+{t}\right)^{{z}} −{y}^{{z}} >{ty}^{{z}−\mathrm{1}} \left(\mathrm{2}+\frac{{t}}{{y}}\right)=\mathrm{2}{ty}^{{z}−\mathrm{1}} +{t}^{\mathrm{2}} {y}^{{z}−\mathrm{2}} \\ $$$${z}\geqslant\mathrm{2}\Rightarrow{y}^{{z}−\mathrm{2}} \geqslant\mathrm{1},\:{y}^{{z}−\mathrm{1}} \geqslant{y} \\ $$$$\Rightarrow{x}^{{z}} −{y}^{{z}} \geqslant\mathrm{2}{ty}+{t}^{\mathrm{2}} ={t}\left(\mathrm{2}{y}+{t}\right)={t}\left({x}+{y}\right) \\ $$$$\Rightarrow\frac{{x}^{{z}} −{y}^{{z}} }{{x}+{y}}\geqslant{t} \\ $$$${if}\:\:\exists{x},{y},{z}\in\mathbb{N},\:{s}.{t}.\:{x}^{{z}} −{y}^{{z}} ={x}+{y},\:{then} \\ $$$$\mathrm{1}\geqslant{t}\in\mathbb{N}\Rightarrow{t}=\mathrm{1} \\ $$$$\Rightarrow{x}={y}+\mathrm{1} \\ $$
Commented by dragan91 last updated on 08/Jul/22
$$\mathrm{well}\:\mathrm{done}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 11/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$