Proof that $t^8+2t^6+4t^4+t^2+1$ is reducible in $\mathbb{F}_p$ Prove that the polynomial $t^8+2t^6+4t^4+t^2+1$ is reducible in $\Bbb F_p$, for all $p\in \Bbb P$.
Here are some examples:
$t^8+2t^6+4t^4+t^2+1=(1 + t + t^4)^2\pmod{2}$
$t^8+2t^6+4t^4+t^2+1=(1 + t) (2 + t) (1 + t^2) (2 + 2 t^2 + t^4)\pmod{3}$
$t^8+2t^6+4t^4+t^2+1=(2 + t^2) (3 + 4 t^2 + t^6)\pmod{5}$
$t^8+2t^6+4t^4+t^2+1=(10 + 9 t + 3 t^2 + 2 t^3 + t^4) (10 + 2 t + 3 t^2 + 9 t^3 + t^4)\pmod{11}$
Thanks in advance.

According to Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$ 
you should calculate galois group of this polynomial over $\mathbb Q$.
If you ask Magma:  

P:=PolynomialAlgebra(Rationals()); f:=x^8+2*x^6+4*x^4+x^2+1; G:=GaloisGroup(f); print G;

then you have

Permutation group G acting on a set of cardinality 8
  Order = 192 = 2^6 * 3
      (1, 2, 3, 4)(5, 6, 7, 8)
      (1, 2)(5, 6)
      (1, 6)(2, 5)
      (1, 6, 7, 8)(2, 3, 4, 5)  

Now you should check that there are no 8-cycles in this group.
Most simple way is to generate all 192 elements and check.
