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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)
forTawa,anoldproblemexplained(1)x+y+z=α(2)x2+y2+z2=β(3)x3+y3+z3=γα,β,γaregiven(4)x4+y4+z4=p(5)x5+y5+z5=qfindp,qwecouldtrytosolvethesystembutit′shardtoexactlysolveitandthencalculatethesumsofthe4thand5thpowers.instead,let′sthinkoftheshapeofthesolutionsofapolynomeof3rddegree.theymightlooklikethis:t1=at2=b−ct3=b+clet′stryit!x=ay=b−cz=b+c⇒(1)a+2b=α(2)a2+2b2+2c=β(3)a3+2b3+6bc=γ(4)a4+2b4+12b2c+2c2−p=0(5)a5+2b5+20b3c+10bc2−q=0doesn′tlookthatbad.wecaneasilysolveequation(1)foraandequation(2)forc(1)a=α−2b(2)c=−a22−b2+β2=[using(1)]−3b2+2αb−α22+β2nowlet′sinsertthesein(3),(4)and(5)(3)−24b3+24αb2−3(3α2−β)b+α3−γ=0(4)−32αb3+32α2b2−4α(3α2−β)b+3α42−α2β+β22−p=0(5)−20(α2+β)b3+20α(α2+β)b2−52(3α4+2α2β−β2)b+α5−q=0dividingbytheconstantfactorsofb2(3)b3−αb2+3α2−β8b−α3−γ24=0(4)b3−αb2+3α2−β8b+2p−3α4+2α2β−β264α=0(5)b3−αb2+3α2−β8b+q−α520(α2+β)=0onlytheredfactorsdiffer⇒theymusthavethesamevalues!⇒−α3−γ24=2p−3α4+2α2β−β264α=q−α520(α2+β)⇒p=α46−α2β+4αγ3+β22q=α56−5α3β6+5α2γ6+5βγ6solvedwithoutsolvingthesameideaistrivialfor2equationsx+y=αx2+y2=βxn+yn=pwithn>2∧n∈Nputx=a−by=a+b⇒2a=α2a2+2b=β⇒a=α2b=β2−α24wemightaswellsolvethesystemandfor4equationsunfortunatelywemustsolvethesystemw+x+y+z=αw2+x2+y2+z2=βw3+x3+y3+z3=γw4+x4+y4+z4=δbutitstillmakesiteasiertosolveputtingw=a−bx=a+by=c−bz=c+d
Commented by Tawa1 last updated on 21/Jul/19
Wow, great, you are helping me more, God will help you more too in no particular order, .Sir Mjs, Sir MrW, sir Tanmay, matabodo and every other person here. God bless everyone here and me too. Hahahaha.
Wow,great,youarehelpingmemore,Godwillhelpyoumoretooinnoparticularorder,.SirMjs,SirMrW,sirTanmay,matabodoandeveryotherpersonhere.Godblesseveryonehereandmetoo.Hahahaha.
Commented by Tawa1 last updated on 25/Jul/19
God bless you sir
Godblessyousir
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 (∗)
letconsiderP(t)=(t−x)(t−y)(t−z)andun=xn+yn+znByreducingP(t)=t3+at2+bt+cwitha=−(x+y+z)b=xy+xz+yzc=−xyzwecaneasilyprovethatun+3+aun+2+bun+1+cun=xnP(x)+ynP(y)+znP(z)=0(∗)Thenu4=−au3−bu2−cu1u5=−au4−bu3−cu2Sowejuztneedtoprovethatabcareknownandfindthema=−α(x+y+z)2=x2+y2+z2+2(xy+xz+yz)⇒b=12(u12−u2)=12(α2−β)(x+y+z)3=(x+y)3+3z(x+y)(x+y+z)+z3=x3+3xy(x+y)+y3+3α(xz+zy)+z3=u3+3xy(α−z)+3α(zx+zy)=u3+3αb−3xyzwegetc=−xyz=13(u13−u3−3αb)=13[α3−γ−3α(α2−β2)]=13[3αβ2−α32−γ]asabcareknownandfoundwecaneasilyfindun∀n>4byusingtheegality(∗)

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