What is $\alpha^{4} + \beta^{4} + \gamma^{4}$?

Question

If $$ \alpha + \beta + \gamma = 14 $$ $$ \alpha^{2} + \beta^{2} + \gamma^{2} = 84 $$ $$ \alpha^{3} + \beta^{3} + \gamma^{3} = 584 $$

What is $\alpha^{4} + \beta^{4} + \gamma^{4}$?

---

Attempt:

Notice that $$ \alpha + \beta + \gamma = 14 \implies \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196$$ $$\alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196 \implies 84 + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196 $$ $$ \alpha \beta + \beta \gamma + \alpha \gamma = 56 $$

Also $$ \alpha + \beta + \gamma = 14 \implies \alpha^{3} + \beta^{3} + \gamma^{3} + 3(\alpha^{2} \beta + \alpha^{2} \gamma + \beta^{2} \alpha + \beta^{2} \gamma + \gamma^{2} \alpha + \gamma^{2} \beta) + 3 \alpha \beta \gamma = 2744$$ $$ 584 = 2744 - 3(\alpha + \beta + \gamma)(\alpha \beta + \alpha \gamma + \beta \gamma) + 6 \alpha \beta \gamma$$ $$ \alpha \beta \gamma = 32$$

So that

$$ (\alpha + \beta + \gamma)^{4} = ( \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) )^{2} $$ $$ 38416 = (\alpha^{2} + \beta^{2} + \gamma^{2})^{2} + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2}) + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)^{2}$$ $$ 38416 = \left[ \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta)^{2} + (\beta \gamma)^{2} + (\alpha \gamma)^{2} \right) \right] + 4 (56)(84) + 4 (56^{2}) $$ $$ 7056 = \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta + \alpha \gamma + \beta \gamma)^{2} - 2\alpha \beta \gamma(\alpha + \beta + \gamma ) \right) $$ $$ 7056= \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( 56^{2} - 2(32)(14) \right) $$ $$\alpha^{4} + \beta^{4} + \gamma^{4} = 2576$$

Answer 1

Well, its visible clearly that numbers are 8, 4 and 2. From first two equations, $$y = 1/2 (-\sqrt{-3 x^2 + 28 x - 28} - x + 14)$$
$$z = 1/2 (\sqrt {-3x^2 + 28 x - 28} - x + 14)$$
Putting them into third equation, we get a cubic equation
$$3x^3 - 42 x^2 + 168 x + 392=584$$ solving which by cardano or any method or writing the symmetric equation for sum and product of roots etc, we see that roots are 2, 4 and 8.

Answer 2

Let $x=\alpha+\beta+\gamma$, $y=\alpha^2+\beta^2+\gamma^2$ and $z=\alpha^3+\beta^3+\gamma^3$ for any real numbers $\alpha,\beta,\gamma$. Then we have the identity: $$\alpha^4+\beta^4+\gamma^4=\frac16\left(x^4-6x^2y+3y^2+8xz\right).\tag{1}$$ You can verify this by expanding everything. In our case, $x=14,y=84$ and $z=584$, so just substitute in the numbers to solve the problem. I get the answer of $4368$.

(Note: I came up with the expression $(1)$ with a computer running Mathematica. It might not be very viable in a competition setting.)

Answer 3

Define $A,B,C$ such that $(x-\alpha)(x-\beta)(x-\gamma) = x^3 - Ax^2 + Bx - C$.

For any $n \ge 1$, let $p_n = \alpha^n + \beta^n + \gamma^n$. In terms of $p_n$, the question becomes:

Given $p_1 = 14, p_2 = 84, p_3 = 584$, what is $p_4$?

There is a bunch of Newton's identities connecting $p_n$ and the coefficients $A,B,C$ above. Using these identities, the computation of $p_4 = \alpha^4 + \beta^4 + \gamma^4$ is mechanical.

• $p_1 = A \\ \quad \implies A = 14$

• $p_2 = A p_1 - 2B \\ \quad \implies B = \frac12(A^2 - p_2) = \frac12(14^2 - 84) = 56$

• $p_3 = A p_2 - B p_1 + 3C \\ \quad \implies C = \frac13 ( p_3 - Ap_2 + B p_1) = \frac13(584 - 14\cdot 84 + 56\cdot 14) = 64$

Finally, we have

$$p^4 = Ap_3 - Bp_2 + Cp_1 = 14\cdot 584 - 56\cdot 84 + 64\cdot 14 = 4368$$

This is different from what you have got. When you compute $\alpha\beta\gamma$, you have make a mistake in one of the coefficient. The correct step should be

$$584 = 2744 - 3(\alpha + \beta + \gamma)(\alpha \beta + \alpha \gamma + \beta \gamma) + \color{red}{3} \alpha \beta \gamma\\ \implies \alpha\beta\gamma = \color{red}{64} $$

Other than this, the rest of your derivation seems fine. In fact, if you substitute $32$ by $64$ in last two lines, you get the right answer. $$\require{cancel} 7056= \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( 56^{2} - 2(\color{red}{\cancelto{64}{\color{gray}{32}}})(14) \right) $$ $$\alpha^{4} + \beta^{4} + \gamma^{4} = \color{red}{\cancelto{4368}{\color{gray}{2576}}} $$