Question Number 82030 by naka3546 last updated on 17/Feb/20
$${a}\:−\:{b}\:+\:{c}\:−\:{d}\:\:=\:\:\mathrm{2} \\ $$$${a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:−\:{d}^{\mathrm{2}} \:\:=\:\:\mathrm{6} \\ $$$${a}^{\mathrm{3}} \:−\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} \:−\:{d}^{\mathrm{3}} \:\:=\:\:\mathrm{20} \\ $$$${a}^{\mathrm{4}} \:−\:{b}^{\mathrm{4}} \:+\:{c}^{\mathrm{4}} \:−\:{d}^{\mathrm{4}} \:\:=\:\:\mathrm{66} \\ $$$${a}\:+\:{b}\:+\:{c}\:+\:{d}\:\:=\:\:? \\ $$
Answered by MJS last updated on 17/Feb/20
$$\mathrm{let} \\ $$$${a}=\alpha−\sqrt{\beta}\wedge{c}=\alpha+\sqrt{\beta}\wedge{b}=\gamma−\sqrt{\delta}\wedge{d}=\gamma+\sqrt{\delta} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{and}\:\mathrm{find} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution} \\ $$$${a}=\mathrm{1}\wedge{b}=\mathrm{0}\wedge{c}=\mathrm{3}\wedge{d}=\mathrm{2} \\ $$$$\Rightarrow\:{a}+{b}+{c}+{d}=\mathrm{6} \\ $$
Commented by mr W last updated on 17/Feb/20
$${perfect}\:{solution}\:{sir}! \\ $$