Prime Powers: Finding $a$ for $a^8+a+1$

[anemone]

Find all positive integers $a$ for which $a^8+a+1$ is a prime number.

 

[topsquark]

Spoiler

\(\displaystyle a^8 + a + 1 = (a^2 + a + 1)(a^6 - a^5 + a^3 - a^2 + 1)\)
(It took me a full page to factor this out. I didn't think of W|A until I checked the answer.)

Since \(\displaystyle a^8 + a + 1\) always factors this way, the only way to get a prime out of it is to set \(\displaystyle a^2 + a + 1 = 1\) and/or \(\displaystyle a^6 - a^5 + a^3 - a^2 + 1 = 1\) and see if the other term is prime. \(\displaystyle a^2 + a + 1 = 1\) for a = 0, 1. \(\displaystyle a^6 - a^5 + a^3 - a^2 + 1 = 1\) for a = 0, (+/-)1. a = +1 is the only solution that gives a prime.

-Dan

 

[kaliprasad]

Find all positive integers $a$ for which $a^8+a+1$ is a prime number.

Spoiler

using a very old trick w (cube root of unity) is a zero

as $w^8 + w + 1 = w^2 + w + 1 = 0$

so $(a-w)$ and $(a-w^2)$ are factors so $a^2 + a + 1$ is a factor

so divdivision
$a^8+a+1 = (a^2+a+1)(a^6-a^5+a^3 -a^2 + 1)$

for a > 1 both are > 1 and so a canot be > 1

for a = 1 we get 3 which is prime so a = 1