Irish Math Olympiad 2009

Find all positive integers \(n\) such that \(n^8+n+1\) is a prime.

Irish Math Olympiad 2009

Solution

Let \(f(x)=x^8+x+1\). If \(w\) is the third root of unity \((w=e^{2i\pi/3})\), then \(w^2+w+1=0\) and \(w^3=1\), so \(w^8=w^2\). Therefore, \(f(w)=w^8+w+1=w^2+w+1=0\). Hence, \(w\) is a root of both \(f(x)=x^8+x+1\) and \(x^2+x+1\), so

\[ f(x)=(x^2+x+1)g(x). \]

We have

\[ x^8+x+1=(x^2+x+1)(x^6-x^5+x^3-x^2+1). \]

Since \(f(1)=1^8+1+1=3\), \(n=1\) gives a prime number.

For \(n>1\),

\[ n^8+n+1=(n^2+n+1)(n^6-n^5+n^3-n^2+1). \]

Both factors are greater than \(1\), so \(n^8+n+1\) is composite for all \(n>1\).

Therefore, the only positive integer satisfying the condition is

\[ \boxed{n=1}. \]

Solution by: Thin Sokkean

Check PDF file to download here