Suppose a^{n+1} + 1 divides a^{2} + a^{2n}, where n is a positive integer. Then a^{n+1} + 1 divides a^{n-1} - a^{2} and hence a^{n+1} + 1 divides a^{4} + 1. Thus in particular n≤3. If n = 3, then a^{n+1} +1 = a^{4} + 1 divides a^{2} + a^{2n} = a^{2} + a^{6}. If n = 1, then a^{2} + 1 divides 2a^{2} implies a^{2} + 1 divides 2, which is not possible. Finally if n = 2, we obtain a^{3} + 1 divides a^{2} + a^{4}, so a^{3} + 1 divides a^{2} - a which again is not possible.
We conclude that if a > 1, then a^{2} + b^{2} is divisible by ab + 1 if and only if n = 3.
f ([0, 1]) | = [0, 1/3]∪[2/3, 1] | |
f (f ([0, 1])) | = [0, 1/9]∪[2/9, 3/9]∪[6/9, 7/9]∪[8/9, 1] | |
f (f (f ([0, 1]))) | = [0, 1/27]∪[2/27, 3/27]∪[6/27, 7, 27]∪[8/27, 9/27]∪ | |
[18/27, 19/27] | ∪[20/27, 21/27]∪[24/27, 25/27]∪[26/27, 1] |
Therefore we may assume that T contains no negative numbers. Now suppose 1/2∈T. Since (x + 2)/3 = 1/2 implies x = - 1/6, we see that 3/2∈T. Now let t_{2} = sup_{t∈T}t and choose t_{3}∈T such that t_{3} > (t_{2} + 2)/3. Since f (T) = T, we see that there exists s∈T such that f (s) > (t_{2} + 2)/3, which is not possible.
We conclude that there is no bounded subset T such that f (T) = T and 1/2∈T.