Irska

Let $f(x)=x^8+x+1$. Numerical values get large very quickly: $$\begin{array}{rl}f(1)& =3\\ f(2)& =259=7\times 37\\ f(3)& =6565=5\times 13\times 101\\ f(4)& =65541=3\times 7\times 3121.\end{array}$$ These numbers may suggest that $f(n)$ will be a prime number only if $n=1$. To prove this, we try to factorise the polynomial $x^8+x+1$. Progress can be made if it is suspected that $x^2+x+1$ is a factor. This can quickly be tested by using a cubic root of unity $\omega \ne 1$. It satisfies ${\omega }^2+\omega +1=0$ and ${\omega }^3=1$, hence ${\omega }^8={\omega }^2$ from which we directly see $f(\omega )=0$. Polynomial division gives now easily the factorisation $f(x)=(x^2+x+1)(x^6-x^5+x^3-x^2+1)$. Another way to obtain this factorisation is the following. We write $x^8+x+1=x^8-x^2+x^2+x+1$ and observe $$x^8-x^2=x^2(x^6-1)=x^2(x^3+1)(x^3-1)=x^2(x^3+1)(x-1)(x^2+x+1).$$ This gives $f(x)=x^8+x+1=(x^2+x+1)(x^2(x^3+1)(x-1)+1)$. If $n\ge 2$, we have $n^2+n+1\ge 7$ and $n^2(n^3+1)(n-1)+1\ge 37$, hence $f(n)$ is not a prime number if $n\ge 2$. As $f(1)=3$ is a prime number, we conclude that $n=1$ is the only positive integer for which $n^8+n+1$ is a prime number.