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Question-173162




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

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