\(n^5-n+2=n\left(n^4-1\right)+2=n\left(n^2+1\right)\left(n^2-1\right)+2\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+2\)

We see: \(\left(n-1\right)\cdot n\cdot\left(n+1\right)⋮3\) (is product of 3 consecutive natural numbers)

\(\Rightarrow n^5-n+2\equiv2\left(mod3\right)\)

But no square numbers are in the form 3k + 2

So \(n\notin\varnothing\)