# Show that no integer of the form $a^3 +1$ is a prime for $a>1$ [duplicate]

Can someone please solve this and explain the steps taken to reach this solution ?

• $x^3+1=(x+1)(x^2-x+1)$ – lulu Jul 27 '16 at 14:28
• In general, you have the following factorization for all odd $n$: $x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y + x^{n-3}y^2-\ldots - xy^{n-2} + y^{n-1})$. It's a nifty thing to remember. – A.Sh Jul 27 '16 at 14:34
• Dudes, this was handled 5 years ago and multiple times in between. In fact back in the day it was deemed only worth a link. Downvotes to all but the honest commenters. – Jyrki Lahtonen Jul 27 '16 at 16:34

Factorise $$a^3 +1 \equiv (a+1)(a^2 -a +1)$$
This is made up of the product of two integers if $a > 1$, namely $a+1$ and $a^2 -a + 1$. i.e: it is the very definition of a composite number. (that is, unless $a+1$ or $a^2 - a + 1$ is equal to $1$, but the hypothesis $a>1$ immediately gives you that.
In fact, it's normally a good idea to factorise $$a^n + b^n = (a+b)(a^{n-1} - a^{n-2}b - \cdots - ab^{n-2} + b^{n-1})$$
Hint: $a^3+1=(a+1)(a^2-a+1)$, so...