1
$\begingroup$

If $$\left\{\begin{array}{c} (\lambda_1 + 1)(\lambda_2 + 1) \cdots (\lambda_n + 1) = 1 \\ (\lambda_1^2 + 1)(\lambda_2^2 + 1) \cdots (\lambda_n^2 + 1) = 1 \\ \vdots \\ (\lambda_1^n + 1)(\lambda_2^n + 1) \cdots (\lambda_n^n + 1) = 1 \end{array}\right.$$ with $\lambda_1, \lambda_2, \ldots, \lambda_n \in \Bbb{C}$, then $$\lambda_1 = \lambda_2 = \cdots = \lambda_n = 0.$$

Sorry I misunderstood the question before, actually it is stated as follows:

If $$ \ (\lambda_1^k + 1)(\lambda_2^k + 1) \cdots (\lambda_n^k + 1) = 1\\ .$$ hold for all positive integers $k$, with $\lambda_1, \lambda_2, \ldots, \lambda_n \in \Bbb{C}$, then $$\lambda_1 = \lambda_2 = \cdots = \lambda_n = 0.$$

$\endgroup$
1
  • $\begingroup$ The question originally asked has been answered below. The revised question is a different matter. I would advise asking the latter in a new post. Make sure to link back to this post, and be sure to include your own thoughts/efforts in solving it, so that it isn't closed this time. $\endgroup$ Aug 23, 2021 at 20:14

1 Answer 1

5
$\begingroup$

In the case of $n=2$ you have $$ a+b+ab=0\\ a^2+b^2+a^2b^2=0 $$ The second equation transforms to $$ 0=(a+b)^2-2ab+a^2b^2=2(a^2b^2-ab). $$ The solution $ab=0$ leads to $a=b=0$. The other solution $ab=1$ gives $a+b=1$, so that $a,b$ are the solutions of the quadratic $x^2-x+1=0$ which has non-zero solutions $\frac{1\pm i\sqrt3}2$.

The claim needs some work.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .