since x2 - 6x + 1 = (x - a)(x - b) = x2 - (a+b)x + ab,
we have a + b = 6, which is clearly an integer. note 5 does not divide 6. note also that ab = 1, as we will need this later.
also (a + b)2 = (a + b)(a + b) - 2ab = 36 - 2 = 34, which is also an integer.
now let's prove an + bn is an integer, by induction on n.
suppose ak-1 + bk-1 is an integer, for all k < n as our induction hypothesis.
then an + bn = (an-1 + bn-1)(a + b) - (ab)(ak-2 + bk-2)
since the RHS side is all integers, the LHS must be, as well.