Question Number 61283 by naka3546 last updated on 01/Jun/19
$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \right)\:\:=\:\:\mathrm{2} \\ $$$$\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)\left({x}^{\mathrm{6}} +{y}^{\mathrm{6}} \right)\left({x}^{\mathrm{8}} \:+\:{y}^{\mathrm{8}} \right)\:\:=\:\:\mathrm{4} \\ $$$$\left({x}^{\mathrm{3}} +\:{y}^{\mathrm{3}} \right)\left({x}^{\mathrm{5}} +\:{y}^{\mathrm{5}} \right)\left({x}^{\mathrm{7}} \:+\:{y}^{\mathrm{7}} \right)\:\:=\:\:\mathrm{6} \\ $$$$\left({x}^{\mathrm{4}} \:+\:{y}^{\mathrm{4}} \right)\left({x}^{\mathrm{5}} \:+\:{y}^{\mathrm{5}} \right)\left({x}^{\mathrm{9}} \:+\:{y}^{\mathrm{9}} \right)\left({x}^{\mathrm{10}} \:+\:{y}^{\mathrm{10}} \right)\:\:=\:\:? \\ $$
Answered by MJS last updated on 02/Jun/19
![let x=p−q, y=p+q with p, q ∈C ⇒ the 1^(st) equation then transforms to q^4 +((4p^2 )/3)q^2 +((4p^6 −1)/(12p^2 ))=0 ⇒ q=±(√(−((2p^2 )/3)±((√(4p^6 +3))/(6p)))) [4 solutions] approximately solving the 3^(rd) equation finally gives the following pairs ((x),(y) ) : (((1.126879)),((−.02685083)) ) (((1.008297+.4838797i)),((1.008297−.4838797i)) ) (((1.116944−.009290636i)),((−.3248961−.8386042i)) ) (((1.116944+.009290636i)),((−.3248961+.8386042i)) ) [of course we can exchange x with y] none of these values satisfy the 2^(nd) equation ⇒ no solution in C](https://www.tinkutara.com/question/Q61394.png)
$$\mathrm{let} \\ $$$${x}={p}−{q},\:{y}={p}+{q}\:\mathrm{with}\:{p},\:{q}\:\in\mathbb{C} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{equation}\:\mathrm{then}\:\mathrm{transforms}\:\mathrm{to} \\ $$$${q}^{\mathrm{4}} +\frac{\mathrm{4}{p}^{\mathrm{2}} }{\mathrm{3}}{q}^{\mathrm{2}} +\frac{\mathrm{4}{p}^{\mathrm{6}} −\mathrm{1}}{\mathrm{12}{p}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:{q}=\pm\sqrt{−\frac{\mathrm{2}{p}^{\mathrm{2}} }{\mathrm{3}}\pm\frac{\sqrt{\mathrm{4}{p}^{\mathrm{6}} +\mathrm{3}}}{\mathrm{6}{p}}}\:\:\:\:\:\left[\mathrm{4}\:\mathrm{solutions}\right] \\ $$$$\mathrm{approximately}\:\mathrm{solving}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{equation} \\ $$$$\mathrm{finally}\:\mathrm{gives}\:\mathrm{the}\:\mathrm{following}\:\mathrm{pairs}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:: \\ $$$$\begin{pmatrix}{\mathrm{1}.\mathrm{126879}}\\{−.\mathrm{02685083}}\end{pmatrix}\:\:\begin{pmatrix}{\mathrm{1}.\mathrm{008297}+.\mathrm{4838797i}}\\{\mathrm{1}.\mathrm{008297}−.\mathrm{4838797i}}\end{pmatrix}\: \\ $$$$\begin{pmatrix}{\mathrm{1}.\mathrm{116944}−.\mathrm{009290636i}}\\{−.\mathrm{3248961}−.\mathrm{8386042i}}\end{pmatrix}\:\:\begin{pmatrix}{\mathrm{1}.\mathrm{116944}+.\mathrm{009290636i}}\\{−.\mathrm{3248961}+.\mathrm{8386042i}}\end{pmatrix} \\ $$$$\left[\mathrm{of}\:\mathrm{course}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exchange}\:{x}\:\mathrm{with}\:{y}\right] \\ $$$$\mathrm{none}\:\mathrm{of}\:\mathrm{these}\:\mathrm{values}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{equation} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{C} \\ $$
Commented by alphaprime last updated on 02/Jun/19
Yes sir absolutely correct said , it got no solutions