α and β are two roots of the equation x² – 6x + 1 = 0

Prove that αⁿ+ βⁿ is an integer for every positive integer value of n, and it is not divisible by 5 for any positive integer value of n

The roots of x² - 6x + 1 = 0 are [3+2*SQRT(2), 3-2*SQRT(2)].

The two expressions for the roots may be given as
[j+k] and [j-k]
where k is the part of the expression that has the radical.
In the binomial expansion of [j+k]ⁿ where n is an even integer, the sign of the operation remains plus (+), and in the binomial expansion of [j-k]ⁿ the sign of the operation alternates between minus (-) where k is raised to an odd power and plus (+) where it is raised to an even power. Thus, when the two exponentiated roots [j+k]ⁿ and [j-k]ⁿ are added, the terms for the odd powers of k (the part of the expression with the radical) cancel each other out. As what remains are the even powers for k, the radical "disappears", leaving a rational number.

The final digit of Aⁿ+ Bⁿ   has a cycle length of 6:
{6, 4, 8, 6, 4, 2}. As a number needs to end in either 0 or 5 to be divisible by 5, and as no 0 or 5 is in the cycle for Aⁿ+ Bⁿ , 
Aⁿ+ Bⁿ can not be divisible by 5.
 

Edited on June 30, 2011, 4:55 pm

Posted by Dej Mar on 2011-06-30 16:51:19