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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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 Another Solution Comment 3 of 3 |
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.

 Posted by Brian Smith on 2017-10-25 13:06:48

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