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Question Number 151849 by mathdanisur last updated on 23/Aug/21

Commented by Rasheed.Sindhi last updated on 23/Aug/21

Special general case  x=y=z(prime of 4m+3 type)  ((z^(2k) +z^(2k) )/(z^k +z^k ))=((2z^(2k) )/(2z^k ))=z^(2k−k) =z^k =z^n ⇒k=n  x=y=z(prime of 4m+3 type) ∧ n=k  For example:  x=y=z=7=4(1)+3 ∧ n=k  x^(2k) +y^(2k) =z^n (x^k +y^k )  ⇒7^(2k) +7^(2k) =7^k (7^k +7^k )  ⇒7^(2k) +7^(2k) =7^(2k) +7^(2k)

$${Special}\:{general}\:{case} \\ $$$${x}={y}={z}\left({prime}\:{of}\:\mathrm{4}{m}+\mathrm{3}\:{type}\right) \\ $$$$\frac{{z}^{\mathrm{2}{k}} +{z}^{\mathrm{2}{k}} }{{z}^{{k}} +{z}^{{k}} }=\frac{\mathrm{2}{z}^{\mathrm{2}{k}} }{\mathrm{2}{z}^{{k}} }={z}^{\mathrm{2}{k}−{k}} ={z}^{{k}} ={z}^{{n}} \Rightarrow{k}={n} \\ $$$${x}={y}={z}\left({prime}\:{of}\:\mathrm{4}{m}+\mathrm{3}\:{type}\right)\:\wedge\:{n}={k} \\ $$$${For}\:{example}: \\ $$$${x}={y}={z}=\mathrm{7}=\mathrm{4}\left(\mathrm{1}\right)+\mathrm{3}\:\wedge\:{n}={k} \\ $$$${x}^{\mathrm{2}{k}} +{y}^{\mathrm{2}{k}} ={z}^{{n}} \left({x}^{{k}} +{y}^{{k}} \right) \\ $$$$\Rightarrow\mathrm{7}^{\mathrm{2}{k}} +\mathrm{7}^{\mathrm{2}{k}} =\mathrm{7}^{{k}} \left(\mathrm{7}^{{k}} +\mathrm{7}^{{k}} \right) \\ $$$$\Rightarrow\mathrm{7}^{\mathrm{2}{k}} +\mathrm{7}^{\mathrm{2}{k}} =\mathrm{7}^{\mathrm{2}{k}} +\mathrm{7}^{\mathrm{2}{k}} \\ $$

Commented by mathdanisur last updated on 23/Aug/21

Thank You Ser

$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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