Hello. The expression can be factored as $(n^4+n^3+n^2+2n+2)(n^4-n^3+n^2-2n+2)$. We now easily observe that $n=1$ is the only solution. $\rule{430pt}{1pt}$ The factorization can be achieved either by creating a difference of squares, i.e. $n^8+n^6+n^4+4=(n^4+n^2+2)^2-(n^3+2n)^2$, or by considering $n^8+n^6+n^4+4=(n^4+an^3+bn^2+cn+2)(n^4+dn^3+en^2+fn+2)$ and solving the system (that's the way I did it in the contest).

how can you be sure that $n^8+n^6+n^4+4=(n^4+an^3+bn^2+cn+2)(n^4+dn^3+en^2+fn+2)$ not $(n^3+an^2+bn+c)(n^5+dn^4+en^3+fn^2+gn+h)$ etc?

gavrilos wrote: The expression can be factored as $(n^4+n^3+n^2+2n+2)(n^4-n^3+n^2-2n+2)$. Since it must be a prime number, one of the factor must be $1$ and since $(n^4+n^3+n^2+2n+2) >(n^4-n^3+n^2-2n+2)$ We get $n^4-n^3+n^2-2n+2=1$ $\Longrightarrow n^4+n^2+1=n^3+2n$ If $n \ge 2$ $\Longrightarrow n^4+n^2+1 > n^3+2n$ Therefore, $n=1$ which is the only solution

Reyan,you can't be sure actually!You also can't be sure that the factors are of $4$-th degree each,as we could have a factor of $2$-nd degree and one of $6$-th degree (there are no other cases since the polynomial has no real roots).When someone is in the contest,they try what seems most reasonable to them.

If don't see this expression; simply let $p=n^8+n^6+n^4+4$; and notice that, $p\mid n^{10}+n^8+n^6+4n^2$. From here, $$ n^{10}+n^8+n^6+4n^2 \equiv n^{10}-n^4-4+4n^2 =n^{10}-(n^2-2)^2 \pmod{p}, $$hence, $p\mid (n^5-n^2+2)(n^5+n^2-2)$. Noticing that for $n\geq 2$; $p$ is much larger than any of those factors; proving that $n=1$ (thus, $p=7$) is the only possibility.