Question Number 64797 by MJS last updated on 21/Jul/19
![for Tawa, an old problem explained (1) x+y+z=α (2) x^2 +y^2 +z^2 =β (3) x^3 +y^3 +z^3 =γ α, β, γ are given (4) x^4 +y^4 +z^4 =p (5) x^5 +y^5 +z^5 =q find p, q we could try to solve the system but it′s hard to exactly solve it and then calculate the sums of the 4^(th) and 5^(th) powers. instead, let′s think of the shape of the solutions of a polynome of 3^(rd) degree. they might look like this: t_1 =a t_2 =b−(√c) t_3 =b+(√c) let′s try it! x=a y=b−(√c) z=b+(√c) ⇒ (1) a+2b=α (2) a^2 +2b^2 +2c=β (3) a^3 +2b^3 +6bc=γ (4) a^4 +2b^4 +12b^2 c+2c^2 −p=0 (5) a^5 +2b^5 +20b^3 c+10bc^2 −q=0 doesn′t look that bad. we can easily solve equation (1) for a and equation (2) for c (1) a=α−2b (2) c=−(a^2 /2)−b^2 +(β/2)=^([using (1)]) −3b^2 +2αb−(α^2 /2)+(β/2) now let′s insert these in (3), (4) and (5) (3) −24b^3 +24αb^2 −3(3α^2 −β)b+α^3 −γ=0 (4) −32αb^3 +32α^2 b^2 −4α(3α^2 −β)b+((3α^4 )/2)−α^2 β+(β^2 /2)−p=0 (5) −20(α^2 +β)b^3 +20α(α^2 +β)b^2 −(5/2)(3α^4 +2α^2 β−β^2 )b+α^5 −q=0 dividing by the constant factors of b^2 (3) b^3 −αb^2 +((3α^2 −β)/8)b−((α^3 −γ)/(24))=0 (4) b^3 −αb^2 +((3α^2 −β)/8)b+((2p−3α^4 +2α^2 β−β^2 )/(64α))=0 (5) b^3 −αb^2 +((3α^2 −β)/8)b+((q−α^5 )/(20(α^2 +β)))=0 only the red factors differ ⇒ they must have the same values! ⇒ −((α^3 −γ)/(24))=((2p−3α^4 +2α^2 β−β^2 )/(64α))=((q−α^5 )/(20(α^2 +β))) ⇒ p=(α^4 /6)−α^2 β+((4αγ)/3)+(β^2 /2) q=(α^5 /6)−((5α^3 β)/6)+((5α^2 γ)/6)+((5βγ)/6) solved without solving the same idea is trivial for 2 equations x+y=α x^2 +y^2 =β x^n +y^n =p with n>2∧n∈N put x=a−(√b) y=a+(√b) ⇒ 2a=α 2a^2 +2b=β ⇒ a=(α/2) b=(β/2)−(α^2 /4) we might as well solve the system and for 4 equations unfortunately we must solve the system w+x+y+z=α w^2 +x^2 +y^2 +z^2 =β w^3 +x^3 +y^3 +z^3 =γ w^4 +x^4 +y^4 +z^4 =δ but it still makes it easier to solve putting w=a−(√b) x=a+(√b) y=c−(√b) z=c+(√d)](Q64797.png)
$$\mathrm{for}\:\mathrm{Tawa},\:\mathrm{an}\:\mathrm{old}\:\mathrm{problem}\:\mathrm{explained} \\ $$$$\left(\mathrm{1}\right)\:\:{x}+{y}+{z}=\alpha \\ $$$$\left(\mathrm{2}\right)\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\beta \\ $$$$\left(\mathrm{3}\right)\:\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\gamma \\ $$$$\alpha,\:\beta,\:\gamma\:\mathrm{are}\:\mathrm{given} \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} ={p} \\ $$$$\left(\mathrm{5}\right)\:\:{x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} ={q} \\ $$$$\mathrm{find}\:{p},\:{q} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{could}\:\mathrm{try}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{hard} \\ $$$$\mathrm{to}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{and}\:\mathrm{then}\:\mathrm{calculate}\:\mathrm{the} \\ $$$$\mathrm{sums}\:\mathrm{of}\:\mathrm{the}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{and}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{powers}. \\ $$$$\mathrm{instead},\:\mathrm{let}'\mathrm{s}\:\mathrm{think}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{solutions}\:\mathrm{of}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree}.\:\mathrm{they} \\ $$$$\mathrm{might}\:\mathrm{look}\:\mathrm{like}\:\mathrm{this}: \\ $$$${t}_{\mathrm{1}} ={a}\:\:{t}_{\mathrm{2}} ={b}−\sqrt{{c}}\:\:{t}_{\mathrm{3}} ={b}+\sqrt{{c}} \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{try}\:\mathrm{it}! \\ $$$$ \\ $$$${x}={a} \\ $$$${y}={b}−\sqrt{{c}} \\ $$$${z}={b}+\sqrt{{c}} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\:{a}+\mathrm{2}{b}=\alpha \\ $$$$\left(\mathrm{2}\right)\:\:{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}=\beta \\ $$$$\left(\mathrm{3}\right)\:\:{a}^{\mathrm{3}} +\mathrm{2}{b}^{\mathrm{3}} +\mathrm{6}{bc}=\gamma \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:\:{a}^{\mathrm{4}} +\mathrm{2}{b}^{\mathrm{4}} +\mathrm{12}{b}^{\mathrm{2}} {c}+\mathrm{2}{c}^{\mathrm{2}} −{p}=\mathrm{0} \\ $$$$\left(\mathrm{5}\right)\:\:{a}^{\mathrm{5}} +\mathrm{2}{b}^{\mathrm{5}} +\mathrm{20}{b}^{\mathrm{3}} {c}+\mathrm{10}{bc}^{\mathrm{2}} −{q}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{doesn}'\mathrm{t}\:\mathrm{look}\:\mathrm{that}\:\mathrm{bad}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{solve} \\ $$$$\mathrm{equation}\:\left(\mathrm{1}\right)\:\mathrm{for}\:{a}\:\mathrm{and}\:\mathrm{equation}\:\left(\mathrm{2}\right)\:\mathrm{for}\:{c} \\ $$$$\left(\mathrm{1}\right)\:\:{a}=\alpha−\mathrm{2}{b} \\ $$$$\left(\mathrm{2}\right)\:\:{c}=−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−{b}^{\mathrm{2}} +\frac{\beta}{\mathrm{2}}\overset{\left[\mathrm{using}\:\left(\mathrm{1}\right)\right]} {=}−\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}\alpha{b}−\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}+\frac{\beta}{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{let}'\mathrm{s}\:\mathrm{insert}\:\mathrm{these}\:\mathrm{in}\:\left(\mathrm{3}\right),\:\left(\mathrm{4}\right)\:\mathrm{and}\:\left(\mathrm{5}\right) \\ $$$$\left(\mathrm{3}\right)\:\:−\mathrm{24}{b}^{\mathrm{3}} +\mathrm{24}\alpha{b}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{3}\alpha^{\mathrm{2}} −\beta\right){b}+\alpha^{\mathrm{3}} −\gamma=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:−\mathrm{32}\alpha{b}^{\mathrm{3}} +\mathrm{32}\alpha^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{4}\alpha\left(\mathrm{3}\alpha^{\mathrm{2}} −\beta\right){b}+\frac{\mathrm{3}\alpha^{\mathrm{4}} }{\mathrm{2}}−\alpha^{\mathrm{2}} \beta+\frac{\beta^{\mathrm{2}} }{\mathrm{2}}−{p}=\mathrm{0} \\ $$$$\left(\mathrm{5}\right)\:\:−\mathrm{20}\left(\alpha^{\mathrm{2}} +\beta\right){b}^{\mathrm{3}} +\mathrm{20}\alpha\left(\alpha^{\mathrm{2}} +\beta\right){b}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{3}\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{2}} \beta−\beta^{\mathrm{2}} \right){b}+\alpha^{\mathrm{5}} −{q}=\mathrm{0} \\ $$$$\mathrm{dividing}\:\mathrm{by}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{factors}\:\mathrm{of}\:{b}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\:{b}^{\mathrm{3}} −\alpha{b}^{\mathrm{2}} +\frac{\mathrm{3}\alpha^{\mathrm{2}} −\beta}{\mathrm{8}}{b}−\frac{\alpha^{\mathrm{3}} −\gamma}{\mathrm{24}}=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:{b}^{\mathrm{3}} −\alpha{b}^{\mathrm{2}} +\frac{\mathrm{3}\alpha^{\mathrm{2}} −\beta}{\mathrm{8}}{b}+\frac{\mathrm{2}{p}−\mathrm{3}\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{2}} \beta−\beta^{\mathrm{2}} }{\mathrm{64}\alpha}=\mathrm{0} \\ $$$$\left(\mathrm{5}\right)\:\:{b}^{\mathrm{3}} −\alpha{b}^{\mathrm{2}} +\frac{\mathrm{3}\alpha^{\mathrm{2}} −\beta}{\mathrm{8}}{b}+\frac{{q}−\alpha^{\mathrm{5}} }{\mathrm{20}\left(\alpha^{\mathrm{2}} +\beta\right)}=\mathrm{0} \\ $$$$\mathrm{only}\:\mathrm{the}\:\mathrm{red}\:\mathrm{factors}\:\mathrm{differ}\:\Rightarrow\:\mathrm{they}\:\mathrm{must}\:\mathrm{have} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{values}! \\ $$$$\Rightarrow\:−\frac{\alpha^{\mathrm{3}} −\gamma}{\mathrm{24}}=\frac{\mathrm{2}{p}−\mathrm{3}\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{2}} \beta−\beta^{\mathrm{2}} }{\mathrm{64}\alpha}=\frac{{q}−\alpha^{\mathrm{5}} }{\mathrm{20}\left(\alpha^{\mathrm{2}} +\beta\right)} \\ $$$$\Rightarrow \\ $$$${p}=\frac{\alpha^{\mathrm{4}} }{\mathrm{6}}−\alpha^{\mathrm{2}} \beta+\frac{\mathrm{4}\alpha\gamma}{\mathrm{3}}+\frac{\beta^{\mathrm{2}} }{\mathrm{2}} \\ $$$${q}=\frac{\alpha^{\mathrm{5}} }{\mathrm{6}}−\frac{\mathrm{5}\alpha^{\mathrm{3}} \beta}{\mathrm{6}}+\frac{\mathrm{5}\alpha^{\mathrm{2}} \gamma}{\mathrm{6}}+\frac{\mathrm{5}\beta\gamma}{\mathrm{6}} \\ $$$$\mathrm{solved}\:\mathrm{without}\:\mathrm{solving} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{idea}\:\mathrm{is}\:\mathrm{trivial}\:\mathrm{for}\:\mathrm{2}\:\mathrm{equations} \\ $$$${x}+{y}=\alpha \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\beta \\ $$$${x}^{{n}} +{y}^{{n}} ={p}\:\mathrm{with}\:{n}>\mathrm{2}\wedge{n}\in\mathbb{N} \\ $$$$\mathrm{put}\:{x}={a}−\sqrt{{b}}\:\:{y}={a}+\sqrt{{b}} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{a}=\alpha \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}=\beta \\ $$$$\Rightarrow\:{a}=\frac{\alpha}{\mathrm{2}}\:\:{b}=\frac{\beta}{\mathrm{2}}−\frac{\alpha^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{we}\:\mathrm{might}\:\mathrm{as}\:\mathrm{well}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{system} \\ $$$$ \\ $$$$\mathrm{and}\:\mathrm{for}\:\mathrm{4}\:\mathrm{equations}\:\mathrm{unfortunately}\:\mathrm{we}\:\boldsymbol{\mathrm{must}} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{system} \\ $$$${w}+{x}+{y}+{z}=\alpha \\ $$$${w}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\beta \\ $$$${w}^{\mathrm{3}} +{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\gamma \\ $$$${w}^{\mathrm{4}} +{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} =\delta \\ $$$$\mathrm{but}\:\mathrm{it}\:\mathrm{still}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{putting} \\ $$$${w}={a}−\sqrt{{b}}\:\:{x}={a}+\sqrt{{b}}\:\:{y}={c}−\sqrt{{b}}\:\:{z}={c}+\sqrt{{d}} \\ $$
Commented by Tawa1 last updated on 21/Jul/19
$$\mathrm{Wow},\:\mathrm{great},\:\mathrm{you}\:\mathrm{are}\:\mathrm{helping}\:\mathrm{me}\:\mathrm{more},\:\mathrm{God}\:\mathrm{will}\:\mathrm{help}\:\mathrm{you}\:\mathrm{more}\:\mathrm{too} \\ $$$$\mathrm{in}\:\mathrm{no}\:\mathrm{particular}\:\mathrm{order}, \\ $$$$.\mathrm{Sir}\:\mathrm{Mjs},\:\:\mathrm{Sir}\:\mathrm{MrW},\:\:\mathrm{sir}\:\mathrm{Tanmay},\:\:\mathrm{matabodo}\:\:\mathrm{and}\:\mathrm{every}\:\mathrm{other}\:\mathrm{person}\:\mathrm{here}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{everyone}\:\mathrm{here}\:\mathrm{and}\:\mathrm{me}\:\mathrm{too}.\:\mathrm{Hahahaha}. \\ $$
Commented by Tawa1 last updated on 25/Jul/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19
![let consider P(t)=(t−x)(t−y)(t−z) and u_n =x^n +y^n +z^n By reducing P (t)=t^3 +at^2 +bt^ +c with a=−(x+y+z) b=xy+xz+yz c=−xyz we can easily prove that u_(n+3) +au_(n+2) +bu_(n+1) +cu_n =x^n P(x)+y^n P(y)+z^n P(z)=0 (∗) Then u_4 =−au_3 −bu_2 −cu_(1 ) u_5 =−au_4 −bu_3 −cu_2 So we juzt need to prove that a b c are known and find them a=−α (x+y+z)^2 =x^2 +y^2 +z^2 +2(xy+xz+yz) ⇒ b=(1/2)(u_1 ^2 −u_2 )=(1/2)(α^2 −β) (x+y+z)^3 = (x+y)^3 +3z(x+y)(x+y+z)+z^3 =x^3 +3xy(x+y)+y^3 +3α(xz+zy)+z^3 =u_3 +3xy(α−z)+3α(zx+zy)=u_3 +3αb−3xyz we get c=−xyz =(1/3)(u_1 ^3 −u_3 −3αb)=(1/3)[α^3 −γ−3α(((α^2 −β)/2))]=(1/3)[((3αβ)/2)−(α^3 /2)−γ] as a b c are known and found we can easily find u_(n ) ∀ n>4 by using the egality (∗)](Q65098.png)
$$ \\ $$$$ \\ $$$$\:\:{let}\:{consider}\:\:{P}\left({t}\right)=\left({t}−{x}\right)\left({t}−{y}\right)\left({t}−{z}\right)\:\:\:{and}\:\:{u}_{{n}} ={x}^{{n}} +{y}^{{n}} +{z}^{{n}} \\ $$$${By}\:\:{reducing}\:{P}\:\left({t}\right)={t}^{\mathrm{3}} +{at}^{\mathrm{2}} +{bt}^{} +{c}\:\:\:\:{with}\:{a}=−\left({x}+{y}+{z}\right)\:\:{b}={xy}+{xz}+{yz}\:\:\:\:\:{c}=−{xyz} \\ $$$${we}\:{can}\:{easily}\:{prove}\:{that}\: \\ $$$${u}_{{n}+\mathrm{3}} +{au}_{{n}+\mathrm{2}} +{bu}_{{n}+\mathrm{1}} +{cu}_{{n}} ={x}^{{n}} {P}\left({x}\right)+{y}^{{n}} {P}\left({y}\right)+{z}^{{n}} {P}\left({z}\right)=\mathrm{0}\:\:\:\left(\ast\right) \\ $$$${Then} \\ $$$${u}_{\mathrm{4}} =−{au}_{\mathrm{3}} −{bu}_{\mathrm{2}} −{cu}_{\mathrm{1}\:\:\:\:} \\ $$$${u}_{\mathrm{5}} =−{au}_{\mathrm{4}} −{bu}_{\mathrm{3}} −{cu}_{\mathrm{2}} \\ $$$${So}\:\:{we}\:{juzt}\:{need}\:{to}\:{prove}\:{that}\:{a}\:{b}\:{c}\:{are}\:{known}\:{and}\:{find}\:{them} \\ $$$${a}=−\alpha\:\:\:\:\:\:\:\:\: \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{xz}+{yz}\right)\:\Rightarrow\:{b}=\frac{\mathrm{1}}{\mathrm{2}}\left({u}_{\mathrm{1}} ^{\mathrm{2}} −{u}_{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\alpha^{\mathrm{2}} −\beta\right) \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{3}} =\:\left({x}+{y}\right)^{\mathrm{3}} +\mathrm{3}{z}\left({x}+{y}\right)\left({x}+{y}+{z}\right)+{z}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{3}} +\mathrm{3}{xy}\left({x}+{y}\right)+{y}^{\mathrm{3}} +\mathrm{3}\alpha\left({xz}+{zy}\right)+{z}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={u}_{\mathrm{3}} +\mathrm{3}{xy}\left(\alpha−{z}\right)+\mathrm{3}\alpha\left({zx}+{zy}\right)={u}_{\mathrm{3}} +\mathrm{3}\alpha{b}−\mathrm{3}{xyz} \\ $$$${we}\:{get}\:\:{c}=−{xyz}\:=\frac{\mathrm{1}}{\mathrm{3}}\left({u}_{\mathrm{1}} ^{\mathrm{3}} −{u}_{\mathrm{3}} −\mathrm{3}\alpha{b}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left[\alpha^{\mathrm{3}} −\gamma−\mathrm{3}\alpha\left(\frac{\alpha^{\mathrm{2}} −\beta}{\mathrm{2}}\right)\right]=\frac{\mathrm{1}}{\mathrm{3}}\left[\frac{\mathrm{3}\alpha\beta}{\mathrm{2}}−\frac{\alpha^{\mathrm{3}} }{\mathrm{2}}−\gamma\right] \\ $$$${as}\:{a}\:\:{b}\:{c}\:{are}\:{known}\:{and}\:{found}\:{we}\:{can}\:{easily}\:{find}\:{u}_{{n}\:} \:\forall\:{n}>\mathrm{4}\:{by}\:{using}\:{the}\:{egality}\:\left(\ast\right) \\ $$