Multiply both sides of x^2 - 6x + 1 = 0 by x^n to form x^(n+2) - 6*x^(n+1) + x^n = 0. Since a and b both satisfy the original quadratic then they both satisfy the new polynomial.
Directly substituting a and b into the new polynomial and adding the resulting expressions yields (a^(n+2) - 6*a^(n+1) + a^n) + (b^(n+2) - 6*b^(n+1) + b^n) = 0 + 0. Which simplifies to [a^(n+2) + b^(n+2)] = 6*[a^(n+1) + b^(n+1)] - [a^n + b^n].
This equation forms a recursive relationship for all the values of a^n+b^n. If n=0 then a^0+b^0 = 2 (trivial). If n=1 then a^1+b^1 = 6 (taken directly from the quadratic).
Then using the recusion n=2 implies a^2+b^2 = 6*6-2 = 34; n=3 implies a^3+b^3 = 6*34-6 = 198; n=4 implies a^4+b^4 = 6*198-34 = 1154; etc. Because the recursion multiples and subtracts integers then all the terms will be integers.
The sequence can be calculated mod 5 to yield 2, 1, 4, 3, 4, 1, 2, 1, 4, 3, etc. This mod 5 sequence repeats with a period of 6 and never includes 0 therefore all terms of the original sequence are coprime to 5.
Another Solution
