How do I find the product $(x_1^2+1)\dotsb(x_{2560}^2+1)$, given that the $x_j$ are roots of $x^{2560} + 2x^{2559} + \dotsb + 2561$?

Question

Let $x_1, x_2,x_3,\dotsc,x_{2560}$ be the roots of the polynomial $$x^{2560}+2x^{2559}+3x^{2558}+\dotsb+2561.$$ What is the value of $$(x_1^2+1)(x_2^2 +1) (x_3^2 +1)\dotsb(x_{2560}^2​+1)?$$

My solution is to use the factorization $$(x^2+1)=(x-i)(x+i).$$

So, let $$P(x)​=x^{2560}+2x^{2559}+3x^{2558}+\dotsb+2561.$$

Then the question becomes $P(i)​P(-i)=\:?$ Am I correct? I'd like to see other solutions.

Answer 1

HINT

If the roots are $\,x_1,\cdots,x_{2560},$ then $P(x)=(x-x_1)\cdots(x-x_{2560}).$
Open the brackets and use Vièta's formulas for coefficients, we have $$\begin{aligned}2=&\,(-1)^1\sum x_k\\3=&\,(-1)^2 \sum_{k\neq j} x_kx_j\\\vdots\\2561=& \;(-1)^{2560}x_1\times \cdots \times x_n \end{aligned}$$ Now, expand independently $$Q^+=(x_1+i)\times \cdots\times (x_{2560}+i)$$ and $$Q^-=(x_1-i)\times \cdots\times (x_{2560}-i)$$ and simplify each of them with the use of the above relations. Then $Q^+\times Q^-$ is the result.