<s>[INST] Is this 5th root in the set of natural numbers? 
Is $$\sqrt[5]{x(x+1)(x^4 + x^2 + 1)} \in \mathbb{N}$$ for some $x$?

I am not asking for all $x$, but just for some natural number $x$?
I don't believe so, but I may be wrong? 
Suppose $x=1$, you get $\sqrt[5]{6}$ which isnt natural. 
$x=2$, you get $\sqrt[5]{126}$
$x=3$, you get: $\sqrt[5]{1092}$
I might just try a proof by induction. Suppose for $x \in \mathbb{N}$:
$$\sqrt[5]{x(x+1)(x^4 + x^2 + 1)} \notin \mathbb{N}$$
But I dont see how that would work out...
 [/INST] Note $(x+1)(x^4+x^2+1)=x^5+x^4+x^3+x^2+x+1$.  But for $x\in\mathbb N, x>0$ we have
$$
x^5 < x^5+x^4+x^3+x^2+x+1 < x^5+5x^4+10x^3+10x^2+5x+1 = (x+1)^5
$$
so that $x^5+x^4+x^3+x^2+x+1$ is not a perfect fifth power, since it it strictly between two consecutive fifth powers.
Next, note that $x$ and $x^5+x^4+x^3+x^2+x+1$ are relatively prime.  If their product is a perfect fifth power, then both of them are perfect fifth powers.  
Thus, $x(x+1)(x^4+x^2+1)$ is not a perfect fifth power.
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