Question Number 211954 by RojaTaniya last updated on 25/Sep/24
$$\:{x},{y}\:{are}\:{rational}\:{numbers}\:{where} \\ $$$$\:{x}\neq\mathrm{0},\:{y}\neq\mathrm{0},\:{x}\neq{y},\:{then}\:{is}\:{it}\: \\ $$$$\:{possible}:\:\:{x}^{\mathrm{5}} +{y}^{\mathrm{5}} =\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:? \\ $$
Answered by Frix last updated on 25/Sep/24
$${x}^{\mathrm{5}} −\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{5}} =\mathrm{0} \\ $$$${y}={px} \\ $$$$\left({p}^{\mathrm{5}} +\mathrm{1}\right){x}^{\mathrm{5}} −\mathrm{2}{p}^{\mathrm{2}} {x}^{\mathrm{4}} =\mathrm{0} \\ $$$$\left({p}^{\mathrm{5}} +\mathrm{1}\right){x}−\mathrm{2}{p}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}{p}^{\mathrm{2}} }{{p}^{\mathrm{5}} +\mathrm{1}}\:\Rightarrow\:{y}=\frac{\mathrm{2}{p}^{\mathrm{3}} }{{p}^{\mathrm{5}} +\mathrm{1}} \\ $$$${p}\in\mathbb{Q}\:\Rightarrow\:\left({x},\:{y}\right)\in\mathbb{Q}^{\mathrm{2}} \\ $$
Commented by RojaTaniya last updated on 25/Sep/24
$$\:{Sir}\:{perfect}. \\ $$