Irish Math Olympiad 2009

 Find all positive integer `n` such that `n^8+n+1` is a prime.

Irish Math Olympiad 2009

Let `f(x)=x^8+x+1` 

If `w` is the `3rd` root of `1` `(w=e^(2ipi/3))`

Then, `w^2+w+1=0` 

            `w^8=w^2` or `8\equiv2(mod3)`

Here, `f(w)=w^8+w+1=w^2+w+1=0`

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

Therefore, `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)`

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

If `n>1` we have `n^8>n^2` or `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)`

So, `n^8+n+1` is composite for all `n>1`.

Solution by: Thin Sokkean

Check PDF file to download here