$\alpha+\beta+\gamma+\delta=0$ and $\alpha^n+\beta^n+\gamma^n+\delta^n=0$ then prove that $\alpha(\alpha+\beta)(\alpha+\gamma)(\alpha+\delta)=0$

Question

Suppose $\alpha,\beta,\gamma,\delta\in\mathbb{R}$ such that $~\alpha+\beta+\gamma+\delta=0~$ and $~\alpha^n+\beta^n+\gamma^n+\delta^n=0~$ (where $n\in\mathbb{N}$ and $n\ne1$),
then prove that $~~\alpha(\alpha+\beta)(\alpha+\gamma)(\alpha+\delta)=0$

The proof is trivial if $n$ is even because we get $\alpha=\beta=\gamma=\delta=0$.
So we consider the case where $n$ is odd.
If $\alpha=0$ then proof is direct so I considered the case where $\alpha\ne0$. Now it is only required to prove that sum of any two of $\alpha,\beta,\gamma,\delta$ should be $0$. The given condition can be simplified as $$\beta^n+\gamma^n+\delta^n=(\beta+\gamma+\delta)^n$$ If $\beta,\gamma,\delta$ is non negative then we can use power-mean inequality to get, $${\beta^n+\gamma^n+\delta^n\over3^n}=\left({\beta+\gamma+\delta\over 3}\right)^n\le{\beta^n+\gamma^n+\delta^n\over 3}$$ $$\implies\alpha^n=\beta^n+\gamma^n+\delta^n=0$$ This contradicts $\alpha\ne0$

But I am not able to prove for the case where at least one of $\beta,\gamma,\delta$ is negative. For example if take $\beta,\delta$ to be positive and $\gamma$ to be negative and then apply power-mean inequality we get $$\left({\beta+\delta\over 2}\right)^n\le{\left(\beta+\gamma+\delta \right)^n+(-\gamma)^n\over2}={\beta^n+\delta^n\over2}$$ What we get is basically the power-mean inequality applied to $\beta$ and $\delta$ and nothing new.

Can someone help me complete the proof or provide an alternative solution. Also could we prove a tighter equality $~(\alpha+\beta)(\alpha+\gamma)(\alpha+\delta)=0~$ to be true (I couldn't find a counter example for it)?

Thanks in advance.

Answer 1

As OP mentioned, the case of $n$ even is obvious. Henceforth, $ n$ is odd.

First, we ignore the special place of $\alpha$ in the final equality and try to understand the implications of $ \alpha + \beta + \gamma + \delta = 0, \alpha^n + \beta^n + \gamma^n + \delta^n = 0 $.
WLOG, let $ \alpha \leq \beta \leq \gamma \leq \delta $.
Let $ a = |\alpha|, b = |\beta|, c = |\gamma|, d = |\delta|$. (This substitution isn't necessary. I strongly prefer dealing with non-negative terms.)
We have the following cases.

Case 1: $ 0 \leq \alpha$.
This requires $a = b = c = d = 0 $.

Case 2: $ \alpha \leq 0 \leq \beta$.
Then $ (b+c+d)^n = a^n = b^n + c^n + d^n$. This requires $ b = c = d = 0$, and hence $ a = 0$.

Case 3: $ \beta \leq 0 \leq \gamma$.
Then $ a + b = c + d = S$ and $ a^n + b^n = c^n + d^n$.
Claim: This requires $a=d, b = c$. (This should feel obvious. If you want to see the rigorous argument, click below.)

Let $ a + b = c + d = 2S$, and $ a = S+P, b = S-P, c = S-R, d = S+R$.
Expanding $ (S+P)^n + (S-P)^n = (S-R)^n + (S+R)^n$, we get $0=\sum{n \choose 2i+1} S^{2i+1} (P^{n-(2i+1) } - R^{n-(2i+1) } )$.
Since $ x^j$ is monotonic. hence equality holds iff $S=0$ or $P = R$.

Case 4: $ \gamma \leq 0 \leq \delta $.
By taking the additive inverse, this is equivalent to case 2.
So $ a = b = c =d = 0 $.

Case 5: $ \delta \leq 0$.
By taking the additive inverse, this is equivalent to case 1.
So $ a = b = c =d = 0 $.

Now, having understood the implications of the condition, we un-ignore the special place of $ \alpha$ in the final equality.

Corollary: We can easily verify that in each of the cases, the original problem statement and OP's tighter equality hypothesis are true:

$$ \alpha (\alpha+\beta)(\alpha+\gamma)(\alpha+\delta) = 0,\\ (\alpha+\beta)(\alpha+\gamma)(\alpha+\delta) = 0.$$

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Note:

• The 3 variable case is an immediate corollary, since we could set any of them to be 0.
• This also shows why the 5 variable case need not hold.

Answer 2

Suppose $\beta\ge \gamma >0>\delta$. Then as $|\delta|=\beta+\gamma$, $\|\delta\|^n=\beta^n+{n\choose 1}\beta^{n-1}\gamma+\cdots +\gamma^n>\beta^n+\gamma^n$, contradiction -- at least when $n>1$ (and for $n=1$, there are counterexamples galore).