# Show that the following non-linear group of equations only have zero solution [closed]

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 $$\lambda_1 = \lambda_2 = \cdots = \lambda_n = 0.$$

• 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. Aug 23, 2021 at 20:14

## 1 Answer

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.