since x^{2}- 6x + 1 = (x - a)(x - b) = x^{2}- (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 a^{n}+ b^{n}is an integer, by induction on n.

suppose a^{k-1}+ b^{k-1}is an integer, for all k < n as our induction hypothesis.

then a^{n}+ b^{n}= (a^{n-1}+ b^{n-1})(a + b) - (ab)(a^{k-2}+ b^{k-2})

since the RHS side is all integers, the LHS must be, as well.