MHB What are the positive integers for which $n^8+n^4+1$ is prime?

[anemone]

Here is this week's POTW:

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Find all positive integers $n$ such that $n^8+n^4+1$ is a prime.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to kaliprasad for his correct solution, which you can find below::)

Spoiler

We have $n^8+n^4+1 = (n^8+2n^4+1)-n^4 = (n^4+1)^2 - (n^2)^2 = (n^4-n^2+1)(n^4+n^2+ 1)$

Now $n^4+n^2+1 > 1$ for positive $n$ and the product shall be prime if:

$n^4-n^2+1 = 1$ and $n^4+n^2+1$ is prime,

which means $n^4 - n^2 = 0$ or $n^2(n^2-1) = n^2(n+1)(n-1) = 0$ and only positive solution is $n =1$

and for $n =1$ we have $n^4+n^2+1=3$ which is prime.

So $n^8+n^4+1$ is prime for only one positive $n$, that is $1$.