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Complex Number and Positive Integer Poser (Posted on 2016-08-27) Difficulty: 3 of 5
Each of X and Y is a complex number such that
A(N) = (XN – YN)/(X-Y) is an integer for 4 consecutive positive integer values of N.

Is A(n) an integer for every positive integer value of N?
If so, prove it.
If not, provide a counter example.

No Solution Yet Submitted by K Sengupta    
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Solution Solution Comment 1 of 1
To solve this problem I started by creating the identities:
A(n+1)*A(n+1) - A(n)*A(n+2) = (x*y)^n
A(n+1)*A(n+2) - A(n)*A(n+3) = (x+y)*(x*y)^n
(x+y)*A(n+1) - (x*y)*A(n) = A(n+2)

We are given four consecutive positive integers make A(n) an integer. Let k, k+1, k+2 and k+3 be those integers.

Using k, k+1, and k+2 then the first identity implies (x*y)^k is an integer.  Similarly, using k+1, k+2, and k+3 then the first identity implies (x*y)^(k+1) is an integer.  Then x*y is a rational number.  For x*y to be rational and have a perfect power that is an integer then x*y must be an integer.

Using k, k+1, k+2, and k+3 in the second identity then implies (x+y)*(x*y)^n is an integer, which then implies x+y is rational.  A(2n) has x+y as a factor: 
A(2n) = (x^(2n)-y^(2n))/(x-y) 
= (x^(2n-2)+x^(2n-4))*y^2+...+x^2*y^(2n-4)+y^(2n-2)) * (x^2-y^2)/(x-y) 
= (x^(2n-2)+x^(2n-4))*y^2+...+x^2*y^(2n-4)+y^(2n-2)) * (x+y)
x+y is a proper factor of A(2n) and if A(2n) is an integer then x+y is also an integer.

x*y and x+y are both integers.  Then A(2) = x+y and A(3) = x^2+x*y+y^2 = (x+y)^2-x*y are both integers.  From this then the third identity implies that A(4) is an integer.  More generally if A(n) and A(n+1) are integers then A(n+2) is an integer, so by induction all A(n) are integers.

I note that with x+y and x*y both integers then x and y are complex conjugate roots of a monic quadratic equation.

Edited on July 16, 2017, 11:18 pm
  Posted by Brian Smith on 2017-07-16 23:16:42

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