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 Sum powers do not take five (Posted on 2011-06-30)
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

 No Solution Yet Submitted by K Sengupta Rating: 2.5000 (2 votes)

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 solution | Comment 1 of 3
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+ B
An+ Bcan not be divisible by 5.

Edited on June 30, 2011, 4:55 pm
 Posted by Dej Mar on 2011-06-30 16:51:19

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