Find Product: $(a^4+b^4+c^4)$ and $(a^6+b^6+c^6)$

[anemone]

Let $a,\,b,\,c$ be numbers such that $(a^2+b^2+c^2)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=26$ and $(a^3+b^3+c^3)\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=78$. Find the value of $(a^4+b^4+c^4)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)$ and $(a^6+b^6+c^6)\left(\dfrac{1}{a^6}+\dfrac{1}{b^6}+\dfrac{1}{c^6}\right)$.

 

[Opalg]

Spoiler: Something goes wrong

Let $S_n = \dfrac{a^n}{b^n} + \dfrac{b^n}{c^n} + \dfrac{c^n}{a^n}$, and let $T_n = \dfrac{b^n}{a^n} + \dfrac{c^n}{b^n} + \dfrac{a^n}{c^n}$. Then $$26 = (a^2+b^2+c^2)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right) = 3 + S_2 + T_2$$ and $$78 = (a^3+b^3+c^3)\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right) = 3 + S_3 + T_3.$$ Therefore $S_2 + T_2 = 23$ and $S_3 + T_3 = 75$.

The equation with roots $\frac ab$, $\frac bc$ and $\frac ca$ is $x^3 - S_1x^2 + T_1x - 1 = 0$, and the equation with roots $\frac ba$, $\frac cb$ and $\frac ac$ is $x^3 - T_1x^2 + S_1x - 1 = 0$. Use Newton's_identities on these equations, to get $$S_2 = S_1^2 - 2T_1,\qquad T_2 = T_1^2 - 2S_1$$. Therefore $$S_2 + T_2 = S_1^2 + T_1^2 - 2(S_1 + T_1) = (S_1+T_1)^2 - 2S_1T_1 - 2(S_1 + T_1).$$ Now write $p=S_1+T_1$ and $q = S_1T_1$, getting $S_2 + T_2 = p^2 - 2p - 2q$, so that $$(*)\quad p^2 - 2p - 2q = 23.$$

In a similar way, using Newton's identities again, $$S_3 = S_1^3 - 3S_1T_1 + 3, \qquad T_3 = T_1^3 - 3S_1T_1 + 3,$$ and so $$S_3 + T_3 = S_1^3 + T_1^3 - 6S_1T_1 + 6 = (S_1 + T_1)^3 - 3S_1T_1(S_1+T_1) - 6S_1T_1 + 6 = p^3 - 3pq - 6q + 6.$$ Therefore $$(**)\quad p^3 - 3pq - 6q = 69.$$

From $(*)$ and $(**)$ it follows that $$p^3 - 3p^2 - 3pq + 6p = 0,$$ $$2p^3 - 6p^2 - 3p(p^2 - 2p - 23) + 12 p = 0,$$ $$-p^3 + 81p = 0.$$ So $p$ is either $0$ or $\pm9$, and the corresponding values for $q$ are $-11.5$, $20$ and $38$. Then $S_1$ and $T_1$, which are the solutions of the equation $x^2 - px + q = 0$, must be the solutions of one of the equations $$x^2 - 11.5 = 0, \qquad x^2 - 9x + 20 = 0, \qquad x^2 + 9x + 38 = 0.$$ The last of those equations has no real solutions. So $\{S_1,T_1\}$ must be either $\{\sqrt{11.5},-\sqrt{11.5}\}$ or $\{4,5\}$. Therefore $\frac ab$, $\frac bc$ and $\frac ca$ (or their reciprocals) are the solutions of one of the equations $
x^3 - \sqrt{11.5}x^2 - \sqrt{11.5}x - 1 = 0$ or $x^3 - 4x^2 + 5x - 1 = 0$. But neither of those cubic equations has three real roots. It follows that there are no real numbers $a,b,c$ satisfying the given equations $(a^2+b^2+c^2)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right) = 26$ and $(a^3+b^3+c^3)\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right) = 78$.

So there is something wrong either with the problem or with my solution.

If complex numbers are allowed then Newton's identities could be used to find values for $(a^4+b^4+c^4)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)$ and $(a^6+b^6+c^6)\left(\dfrac{1}{a^6}+\dfrac{1}{b^6}+\dfrac{1}{c^6}\right)$. But then there would be three possible values for $\{S_1,T_1\}$, and the problem would not have a unique solution.

 

[anemone]

Hi Opalg! Thanks for participating in this unsolved challenge. I just double checked your solution and it seems there is nothing wrong with it. But just to let you know, I believe this is a competition problem from a high school in China and I will wait for the poster to mention anything about the validity of the problem and post back here.

 

[Theia]

Spoiler

I remember, I put this into my CAS and it was just boring algebraic equation manipulation. Such boring that I didn't even post the solution to give someone else the happiness... Unfortunately I don't have much time now to reproduce my solution. Maybe someone could give this another shot?