Question Number 175108 by Mastermind last updated on 19/Aug/22
Answered by MJS_new last updated on 19/Aug/22
$${y}={px} \\ $$$$\begin{cases}{\left({p}^{\mathrm{4}} +\mathrm{6}{p}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{4}} =\mathrm{1}}\\{\left({p}+\mathrm{1}\right)^{\mathrm{4}} {x}^{\mathrm{5}} ={x}\left(\mathrm{1}−{p}\right)}\end{cases} \\ $$$${x}=\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{y} \\ $$$$\begin{cases}{{x}^{\mathrm{4}} =\frac{\mathrm{1}}{{p}^{\mathrm{4}} +\mathrm{6}{p}^{\mathrm{2}} +\mathrm{1}}}\\{{x}^{\mathrm{4}} =\frac{\mathrm{1}−{p}}{\left({p}+\mathrm{1}\right)^{\mathrm{4}} }}\end{cases} \\ $$$$\frac{\mathrm{1}}{{p}^{\mathrm{4}} +\mathrm{6}{p}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}−{p}}{\left({p}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$${p}\left({p}^{\mathrm{4}} +\mathrm{10}{p}^{\mathrm{2}} +\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{p}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{0}\:\Rightarrow\:{x}=\pm\mathrm{1} \\ $$