Five roots of $x^5+x+1=0$ and the value of $\prod_{k=1}^{5} (2+x_k^2)$

Hint: $x^2+2=(x+\sqrt{2}i)(x-\sqrt{2}i)$

$x^5+x+1= (x-x_1)(x-x_2)\cdot...\cdot(x-x_5) \tag 1$

now put $x=\sqrt{2}i$, $x=-\sqrt{2}i$ in $(1)$ and multiply both equations.

Let us transform $x^5+x+1=0$, by $y=2+x^2 \implies x=(y-2)^{1/2}$, Then we get $$(y-2)^{5/2}+(y-2)^{1/2}=-1$$ sqyarinf this w have $$(y-2)^5+(y-2)+2(y-2)^3-1=0$$ The required expression is nothing but the product of roots of this $y$-equation namely $y_1 y_2y_3y_4y_5$ Hence $$\prod_{k=1}^{5} (2+x_k^2)=y_1 y_2y_3y_4,y_5= -[-32-2 +2(-8)-1]= 51.$$

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Note that $\prod_{k=1}^5 (2+x_k^2) = \prod_{k=1}^5 (\sqrt{2}+ix_k)(\sqrt 2 - ix_k)$.

This is the product of all roots of a polynomial, whose roots are exactly $\sqrt{2} \pm ix_k$ for $k=1,...,5$.

Note that if $x^5+x+1$ has roots $x_1,...,x_5$, then $p(y) = (-iy+\sqrt 2i)^5 + (-iy+\sqrt 2i) + 1$ has roots $\sqrt 2 + ix_k$, $k=1,...,5$. The conjugate of this polynomial $\bar{p}$ has roots $\sqrt 2 - ix_k$.

Which means that the polynomial which has roots exactly equal to those we want, is $p\bar p$, and we need just the constant term of this whole polynomial, because by Vieta that is the product of all the roots. The constant of $p$ is $\sqrt 2^5i^5 + \sqrt 2i^5 +1 = 5\sqrt 2i + 1$, similarly of $\bar{p}$ is $1-5\sqrt 2 i$. Multiply these to get $1+(5\sqrt 2)^2 = 1+50=51$ and we are done.