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