Math Book Cambodia: Irish Math Olympiad 2009

Sunday, March 14, 2021

Find all positive integer $n$ $n$ such that $n^{8} + n + 1$ $n^8+n+1$ is a prime.

Irish Math Olympiad 2009

Let $f (x) = x^{8} + x + 1$ $f(x)=x^8+x+1$

If $w$ $w$ is the $3 r d$ $3rd$ root of $1$ $1$ $(w = e^{2 i \frac{π}{3}})$ $(w=e^(2ipi/3))$

Then, $w^{2} + w + 1 = 0$ $w^2+w+1=0$

$w^{8} = w^{2}$ $w^8=w^2$ or $8 \equiv 2 (mod 3)$ $8\equiv2(mod3)$

Here, $f (w) = w^{8} + w + 1 = w^{2} + w + 1 = 0$ $f(w)=w^8+w+1=w^2+w+1=0$

Then, $w$ $w$ is a root of $f (x) = x^{8} + x + 1$ $f(x)=x^8+x+1$ and $x^{2} + x + 1$ $x^2+x+1$

Therefore, $f (x) = (x^{2} + x + 1) g (x)$ $f(x)=(x^2+x+1)g(x)$

We can see that: $f (x) = x^{8} + x + 1 = (x^{2} + x + 1) (x^{6} - x^{5} + x^{3} - x^{2} + 1)$ $f(x)=x^8+x+1=(x^2+x+1)(x^6-x^5+x^3-x^2+1)$

Note, $f (1) = 3$ $f(1)=3$ is a prime.

If $n > 1$ $n>1$ we have $n^{8} > n^{2}$ $n^8>n^2$ or $n^{8} + n + 1 > n^{2} + n + 1 > 1$ $n^8+n+1>n^2+n+1>1$

$n^{8} + n + 1 = (n^{2} + n + 1) (n^{6} - n^{5} + n^{3} - n^{2} + 1)$ $n^8+n+1=(n^2+n+1)(n^6-n^5+n^3-n^2+1)$

So, $n^{8} + n + 1$ $n^8+n+1$ is composite for all $n > 1$ $n>1$ .

Solution by: Thin Sokkean

Check PDF file to download here

Math Book Cambodia