If $a+b+c=0$ prove that $3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4)$

If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove

$$3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4)$$

I made this question as a more difficult (higher degree) version of this question. My idea was that algebraic brute force methods are easy to distinguish from more sophisticated ones if the degree of the terms in the question is higher. The question was specifically made using the method from my answer to the linked question.

• set $$c=-a-b$$ this will solve the problem – Dr. Sonnhard Graubner Oct 12 '17 at 20:58
• Maybe it's a linear combination of $(a+b+c)^n = 0$ for $n = 1, 2, \ldots, 7$? (Of course, maybe not, if it also requires terms like $(ab+ac+bc)(a+b+c)^3 = 0$ to show the difference is in the ideal generated by $a+b+c$.) – Daniel Schepler Oct 12 '17 at 21:01

3 Answers

Let's define $$S_n=a^n+b^n+c^n$$ and consider the generating function $$F(t)=\sum_{n=0}^\infty S_nt^n=\frac1{1-at}+\frac1{1-bt}+\frac1{1-ct}.$$ Using $a+b+c=0$ gives $$(1-at)(1-bt)(1-ct)=1+pt^2+qt^3$$ for some $p$ and $q$, and $$F(t)=\frac{3+pt^2}{1+pt^2+qt^3}.$$ From this we expand as a power series $$F(t)=(3+pt^2)(1-(pt^2+qt^3)+(pt^2+qt^3)^2-\cdots) =3-2pt^2-3qt^3+2p^2t^4+5pqt^5+\cdots$$ and now you can read off this, and several other identities...

• That's a great method! – Orest Bucicovschi Oct 12 '17 at 22:06
• @orangeskid Really? Try to end the starting problem by this method and you'll understand that practically my solution is much more better. See please my solution. – Michael Rozenberg Oct 13 '17 at 6:05

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, $u=0$ and $$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=-6v^2,$$ $$a^3+b^3+c^3=a^3+b^3+c^3-3abc+3abc=$$ $$=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc=3abc=3w^3,$$ $$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)=$$ $$=36v^4-2((ab+ac+bc)^2-2abc(a+b+c))=36v^4-18v^4=18v^4$$ and $$a^5+b^5+c^5=(a^3+b^3+c^3)(a^2+b^2+c^2)-\sum_{cyc}(a^3b^2+a^3c^2)=$$ $$=(-6v^2)\cdot3w^3-(a+b+c)(a^2b^2+a^2c^2+b^2c^2)+abc(ab+ac+bc)=$$ $$=-18v^2w^3+3v^2w^3=-15v^2w^3.$$ Id est, $$3(a^2+b^2+c^2)(a^5+b^5+c^5)-5(a^3+b^3+c^3)(a^4+b^4+c^4)=$$ $$=3(-6v^2)(-15v^2w^3)-5\cdot3w^3\cdot18v^4=0.$$ Done!

• @Peter Sheldrick Firstly, you need to make it. And other is: you need to get the above formula. All these take a time. – Michael Rozenberg Oct 13 '17 at 14:14

plug in the term $$c=-a-b$$ in the left-hand side of the equation we get $$-30 a b (a+b) \left(a^2+a b+b^2\right)^2$$ and so is the right-hand side a remark: if we compute the left-hand side minus the right-hand side and factorize, we obtain $$-(a+b+c) \left(2 a^6-2 a^5 b-2 a^5 c-a^4 b^2+4 a^4 b c-a^4 c^2+6 a^3 b^3-3 a^3 b^2 c-3 a^3 b c^2+6 a^3 c^3-a^2 b^4-3 a^2 b^3 c+6 a^2 b^2 c^2-3 a^2 b c^3-a^2 c^4-2 a b^5+4 a b^4 c-3 a b^3 c^2-3 a b^2 c^3+4 a b c^4-2 a c^5+2 b^6-2 b^5 c-b^4 c^2+6 b^3 c^3-b^2 c^4-2 b c^5+2 c^6\right)$$ and it is clear that we get zero