Is it true? $(1+ta_1)(1+ta_2)\cdots(1+ta_n)=1, \;\forall t\in{\Bbb R}$ ⟺ $a_1=a_2=...=a_n=0$

Question

$(1+ta_1)(1+ta_2)...(1+ta_n)=1$ $\forall t\in R$ if and only if $a_1=a_2=...=a_n=0$

Is it true? If yes, how to prove it?

I already gave a proof when $n=3$.

$1=(1+ta_1)(1+ta_2)(1+ta_3)=1+(a_1+a_2+a_3)t+(a_1a_2+a_1a_3+a_2a_3)t^2+(a_1a_2a_3)t^3$

It is equivalent to

$a_1+a_2+a_3=0$

$a_1a_2+a_1a_3+a_2a_3=0$

$a_1a_2a_3=0$

And this gives

$a_1=a_2=a_3=0$

Is it possible to generalize it for arbitrary $n$?

$(1+ta_1)(1+ta_2)\cdots(1+ta_n)=1$, $\forall{t}$ $\in$ $\Bbb{R}$ ⟺ $a_1=a_2=\cdots=a_n=0$

Answer 1

If there exist an $\,a_k \ne 0\,$ then for $\,t = -1/a_k\,$ the LHS is $\,0\,$ i.e. different from the RHS which is $\,1\,$. Therefore the equality can hold for all real $\,t\,$ iff $\,a_k = 0 \mid k=1,2,\ldots,n\,$.

Note: the above assumes that the coefficients $\,a_k\,$ are real.

Answer 2

I've just found a new way to prove this.

It's not hard to see that the coefficients of polynomial are all zeros. When $n=4$, that is to say,

$a_1+a_2+a_3+a_4=0$

$a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4=0$

$a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4=0$

$a_1a_2a_3a_4=0$

By the last equality, at least one $a_i(i=1,2,3,4)$ is zero. Let's assume $a_1=0$.

Then by the second last equality, $a_2a_3a_4=0$. Let's assume $a_2=0$.

Then by the third last equality, $a_3a_4=0$. Let's assume $a_3=0$.

Then by the first equality, $a_4=0$.