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Prove that there do not exist $n>1$ complex numbers $z_1, z_2, \ldots, z_n$, no two equal, such that for all $1 \le k \le n$

$$ \prod\limits_{i\neq k} (z_k-z_i)=\prod\limits_{i\neq k} (z_k+z_i)$$

At the first look it seems too easy to solve but after trying some methods I can't solve it.It seems to have no solutions so I tried to show their difference can't always equal to zero but I can't.The case $n=2$ Could easily solved by $$z_1-z_2 = z_1+z_2 = z_2+z_1 = z_2-z_1$$

Which forces to have $z_1=z_2$ The case $n=3$ takes some time but it is not very hard to check we don't have solutions there too.But I don't know how to solve it in general case maybe induction could work but I can't solve it using that too.Any hints?

Source:Iran third round math olympiad.

Taha Akbari
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2 Answers2

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First reduce to the case where all of the $z_i$ are non-zero.

Define polynomials $p,q,r$ by $p(z)=\prod_{i=1}^n(z-z_i)$ and $q(z)=2z\frac{dp}{dz}$ and $r(z)=\prod_{i=1}^n(z+z_i).$ Then your equations are equivalent to $q(z_k)=r(z_k)$ for $1\leq k\leq n.$ This implies $q+(-1)^np=r:$ both sides match at $n$ points and at zero. But then the leading coefficients only match if $2n+(-1)^n=1,$ which forces $n\leq 1.$

Dap
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  • Could you explain how you get the result $ q(z_k)=r(z_k)$ for $1≤k≤n$, please? I tried it for the case $k=2$, but it didn't work out. – John Bentin Nov 17 '17 at 12:38
  • @JohnBentin: $dp/dz$ is $\sum_{k=1}^n\prod_{i\neq k}(z-z_i)$ (I don't actually need that it's a derivative). So $q(z_k)=2z_k\prod\limits_{i\neq k} (z_k-z_i)=2z_k\prod\limits_{i\neq k} (z_k+z_i)=r(z_k).$ – Dap Nov 17 '17 at 15:47
  • So, to get this (false) result, you are assuming true the equality that is to be proved impossible. I think that you should make clear that you are using this method of proof in the statement of the proof. – John Bentin Nov 17 '17 at 16:32
  • @JohnBentin: fair enough – Dap Nov 17 '17 at 18:09
  • What if $z_i=0$ for some $i$? – Taha Akbari Nov 17 '17 at 20:07
  • @TahaAkbari: the first step is to reduce to the case where they're all non-zero. You did ask for hints, so I didn't give all the steps. You can basically just delete that element from the list $z_1,\dots,z_n,$ at least for $n\geq 3.$ – Dap Nov 18 '17 at 08:25
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Proof by contradiction. Suppose such a set exists.

Let $f(z ) = \prod (z - z_i)$.
Consider $f'(z) = \sum \prod_{i \neq k} (z-z_i)$
$f'(z_k) = \prod_{i\neq k} (z_k-z_i)$.

Similarly, let $g(z) = \prod (z+z_i)$
Consider $g'(z) = \sum \prod_{i \neq k} (z+z_i)$
$g'(z_k) = \prod_{i\neq k} (z_k+z_i)$.

Since $f'(z) - g'(z) = 0$ on the $n$ values $z_k$ and has degree at most $n-1$, hence $f'(z) - g'(z) \equiv 0$.
Integrating, we get $f(z) \equiv g(z)$, so we can pair up the roots $\{z_i \} = \{-z_i\}$, except possibly for 0.

Since $n > 1$, WLOG let $0 \neq z_1 = -z_2$. Then, we get the contradiction

$$ 0 \neq \prod_{i\neq 1} (z_1 - z_i) = \prod_{i\neq 1 } (z_1 + z_i) = (z_1 + z_2) \prod_{i\neq 1,2 } (z_1 + z_i) = 0. $$

While this solution was really natural to construct, it's likely not the official solution because it used calculus (integrating) in an olympiad problem.

Calvin Lin
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