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Question Number 90793 by kelum last updated on 26/Apr/20
a+b+c+d=4 a^2 +b^2 +c^2 +d^2 =10 a^3 +b^3 +c^3 +d^3 =22 a^4 +b^4 +c^4 +d^4 = ?
$${a}+{b}+{c}+{d}=\mathrm{4} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{10} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} =\mathrm{22} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} =\:? \\ $$
Commented by mr W last updated on 26/Apr/20
you are sure the question is correct? we have four unknowns but only three eqn., so we can not get an unique solution.
$${you}\:{are}\:{sure}\:{the}\:{question}\:{is}\:{correct}? \\ $$$${we}\:{have}\:{four}\:{unknowns}\:{but}\:{only} \\ $$$${three}\:{eqn}.,\:{so}\:{we}\:{can}\:{not}\:{get}\:{an}\:{unique} \\ $$$${solution}. \\ $$
Commented by kelum last updated on 26/Apr/20
Yes its correct. I know the answers Just dont know how to get tbe answers.
$${Yes}\:{its}\:{correct}.\:{I}\:{know}\:{the}\:{answers} \\ $$$${Just}\:{dont}\:{know}\:{how}\:{to}\:{get}\:{tbe}\: \\ $$$${answers}. \\ $$$$ \\ $$
Commented by mr W last updated on 26/Apr/20
what is the answer? i got no unique solution. we can only say a^4 +b^4 +c^4 +d^4 ≥49
$${what}\:{is}\:{the}\:{answer}? \\ $$$${i}\:{got}\:{no}\:{unique}\:{solution}. \\ $$$${we}\:{can}\:{only}\:{say} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} \geqslant\mathrm{49} \\ $$
Commented by mr W last updated on 29/Apr/20
if you know the answer, please share it!
$${if}\:{you}\:{know}\:{the}\:{answer},\:{please}\:{share} \\ $$$${it}! \\ $$
Answered by mr W last updated on 26/Apr/20
p_1 =a+b+c=4−d p_2 =a^2 +b^2 +c^2 =10−d^2 p_3 =a^3 +b^3 +c^3 =22−d^3 ⇒e_1 =4−d p_2 =e_1 ^2 −2e_2 ⇒10−d^2 =(4−d)^2 −2e_2 ⇒e_2 =d^2 −4d+3 p_3 =e_1 ^3 −3e_1 e_2 +3e_3 ⇒22−d^3 =(4−d)^3 −3(4−d)(d^2 −4d+3)+3e_3 ⇒22−d^3 =(4−d)(−2d^2 +4d+7)+3e_3 ⇒e_3 =−d^3 +4d^2 −3d−2 p_4 =e_1 ^4 −4e_1 ^2 e_2 +4e_1 e_3 +2e_2 ^2 −4e_4 p_4 =(4−d)^4 −4(4−d)^2 (d^2 −4d+3)+4(4−d)(−d^3 +4d^2 −3d−2)+2(d^2 −4d+3)^2 p_4 =3d^4 −16d^3 +12d^2 +8d+50 a^4 +b^4 +c^4 =3d^4 −16d^3 +12d^2 +8d+50 a^4 +b^4 +c^4 +d^4 =4d(d−2)(d^2 −2d−1)+50≠constant
$${p}_{\mathrm{1}} ={a}+{b}+{c}=\mathrm{4}−{d} \\ $$$${p}_{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{10}−{d}^{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{22}−{d}^{\mathrm{3}} \\ $$$$\Rightarrow{e}_{\mathrm{1}} =\mathrm{4}−{d} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \: \\ $$$$\:\Rightarrow\mathrm{10}−{d}^{\mathrm{2}} =\left(\mathrm{4}−{d}\right)^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \\ $$$$\:\Rightarrow{e}_{\mathrm{2}} ={d}^{\mathrm{2}} −\mathrm{4}{d}+\mathrm{3} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}{e}_{\mathrm{1}} {e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{22}−{d}^{\mathrm{3}} =\left(\mathrm{4}−{d}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{4}−{d}\right)\left({d}^{\mathrm{2}} −\mathrm{4}{d}+\mathrm{3}\right)+\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{22}−{d}^{\mathrm{3}} =\left(\mathrm{4}−{d}\right)\left(−\mathrm{2}{d}^{\mathrm{2}} +\mathrm{4}{d}+\mathrm{7}\right)+\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow{e}_{\mathrm{3}} =−{d}^{\mathrm{3}} +\mathrm{4}{d}^{\mathrm{2}} −\mathrm{3}{d}−\mathrm{2} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} ^{\mathrm{4}} −\mathrm{4}{e}_{\mathrm{1}} ^{\mathrm{2}} {e}_{\mathrm{2}} +\mathrm{4}{e}_{\mathrm{1}} {e}_{\mathrm{3}} +\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{e}_{\mathrm{4}} \\ $$$${p}_{\mathrm{4}} =\left(\mathrm{4}−{d}\right)^{\mathrm{4}} −\mathrm{4}\left(\mathrm{4}−{d}\right)^{\mathrm{2}} \left({d}^{\mathrm{2}} −\mathrm{4}{d}+\mathrm{3}\right)+\mathrm{4}\left(\mathrm{4}−{d}\right)\left(−{d}^{\mathrm{3}} +\mathrm{4}{d}^{\mathrm{2}} −\mathrm{3}{d}−\mathrm{2}\right)+\mathrm{2}\left({d}^{\mathrm{2}} −\mathrm{4}{d}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$${p}_{\mathrm{4}} =\mathrm{3}{d}^{\mathrm{4}} −\mathrm{16}{d}^{\mathrm{3}} +\mathrm{12}{d}^{\mathrm{2}} +\mathrm{8}{d}+\mathrm{50} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{3}{d}^{\mathrm{4}} −\mathrm{16}{d}^{\mathrm{3}} +\mathrm{12}{d}^{\mathrm{2}} +\mathrm{8}{d}+\mathrm{50} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} =\mathrm{4}{d}\left({d}−\mathrm{2}\right)\left({d}^{\mathrm{2}} −\mathrm{2}{d}−\mathrm{1}\right)+\mathrm{50}\neq{constant} \\ $$
Commented by mr W last updated on 26/Apr/20
or p_1 =a+b+c+d=4 p_2 =a^2 +b^2 +c^2 +d^2 =10 p_3 =a^3 +b^3 +c^3 +d^3 =22 p_1 =e_1 =4 p_2 =e_1 ^2 −2e_2 ⇒10=4^2 −2e_2 ⇒e_2 =3 p_3 =e_1 ^3 −3e_1 e_2 +3e_3 ⇒22=64−36+3e_3 ⇒e_3 =−2 p_4 =e_1 ^4 −4e_1 ^2 e_2 +4e_1 e_3 +2e_2 ^2 −4e_4 =4^4 −4×4^2 ×3+4×4×(−2)+2×3^2 −4e_4 =50−4abcd ⇒a^4 +b^4 +c^4 +d^4 =50−4abcd≠constant
$${or} \\ $$$${p}_{\mathrm{1}} ={a}+{b}+{c}+{d}=\mathrm{4} \\ $$$${p}_{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{10} \\ $$$${p}_{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} =\mathrm{22} \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{4} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \:\Rightarrow\mathrm{10}=\mathrm{4}^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \:\Rightarrow{e}_{\mathrm{2}} =\mathrm{3} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}{e}_{\mathrm{1}} {e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \:\Rightarrow\mathrm{22}=\mathrm{64}−\mathrm{36}+\mathrm{3}{e}_{\mathrm{3}} \:\Rightarrow{e}_{\mathrm{3}} =−\mathrm{2} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} ^{\mathrm{4}} −\mathrm{4}{e}_{\mathrm{1}} ^{\mathrm{2}} {e}_{\mathrm{2}} +\mathrm{4}{e}_{\mathrm{1}} {e}_{\mathrm{3}} +\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{e}_{\mathrm{4}} \\ $$$$=\mathrm{4}^{\mathrm{4}} −\mathrm{4}×\mathrm{4}^{\mathrm{2}} ×\mathrm{3}+\mathrm{4}×\mathrm{4}×\left(−\mathrm{2}\right)+\mathrm{2}×\mathrm{3}^{\mathrm{2}} −\mathrm{4}{e}_{\mathrm{4}} \\ $$$$=\mathrm{50}−\mathrm{4}{abcd} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} =\mathrm{50}−\mathrm{4}{abcd}\neq{constant} \\ $$

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