show that x^10+x^5+1 cannnot be a prime for x >1 ( x is integer)?

Answer 1

A consultation with PARI shows that x²+x+1
is a factor of x^10+x^5+1.
In fact, x^10+x^5+1 = (x²+x+1)(x^8 - x^7 + x^5 -x^4 + x^3 -x+1).
So no matter what n you put in here it will always
be divisible by n²+n+1.
Examples: 2^10 + 2^5+1 = 1024+33 = 1057
is divisible by 7.
Let's check out 3:
3^10 + 3^5+1 = 59293 = 13*4561.
If n>1, n²+n1 will always be a proper divisor of n^10 + n^5 +1,
so your result follows.
Lest you think the factorisation given above is
"pulling a rabbit out of a hat", it can be proven
without PARI.
Let's factor x^15 -1 in two different ways.
First, it is the difference of 2 cubes:
x^15 - 1 = (x^5-1)(x^10 + x^5 + 1)
On the other hand,
x^15 -1 = Φ_1 Φ_3 Φ_5 Φ_15,
where Φ_i is the cyclotomic polynomial of degree i.
Now Φ_1Φ_5 = x^5-1.
So x^10 + x^5 + 1
= Φ_3 Φ15 = (x²+x+1)(x^8 - x^7 + x^5 -x^4 + x^3 -x+1).
One can use this same idea to show that
if n is not divisible by 3
x^2n + x +1 is always divisible by x²+x+1.