Factoring $a^{10}+a^5+1$ I'm very interested to know how I can factorise $a^{10} +a^5 +1$ in two factors with integer coefficients. I've tried a lot but I don't have any idea how do that.

Another hint: can you factor $a^{15}-1$? In more than one way?
Six years later, for the record, let's spell out what I meant by this.  What I had intended was to observe
$$a^{15}-1 = (a^3)^5 - 1 = (a^3-1)(a^{12} + a^9 + a^6 + a^3 + 1) 
= (a-1)(a^2+a+1)(a^{12} + a^9 + a^6 + a^3 + 1)$$
and also
$$a^{15}-1 = (a^5)^3 - 1 = (a^5-1)(a^{10} + a^5 + 1) = (a-1) (a^4+a^3+a^2+a+1) (a^{10}+a^5+1)$$
Thus we have
$$(a^2+a+1) (a^{12}+a^9 + a^6 + a^3 + 1) = (a^4 + a^3 + a^2 + a + 1)(a^{10} + a^5 + 1)$$
and so
$$a^{10} + a^5 + 1 = (a^2 + a + 1) \times {a^{12} + a^9 + a^6 + a^3 + 1 \over a^4 + a^3 + a^2 + a + 1}$$
Now, since $a^2 + a + 1 = (a^3-1)/(a-1)$, the roots of $a^2 + a + 1$ are just the two complex cube roots of unity; it's easy to check that $a^{10} + a^5 + 1$ has those same roots. So that quotient is in fact a polynomial.  We can find out what polynomial it is by long division, and we get the answer:
$$a^{10} + a^5 + 1 = (a^2 + a + 1) (a^8 - a^7 + a^5 - a^4 + a^3 - a + 1).$$
