Consider a sequence generated by positive integer x where each term is of the form ⌊(2+√7)x⌋. Prove that the sum of two consecutive terms will always be odd.
First, I will analyze (2+sqrt(7))^x + (2-sqrt(7))^x.
Notice that when the binomials are expanded that any term with a sqrt(7) raised to an odd power is positive in the first and negative in the second expansion.
Then all these terms cancel leaving two copies of every integer term. This them means that (2+sqrt(7))^x + (2-sqrt(7))^x is an even integer for all x.
0 > 2-sqrt(7) > -1, then (2-sqrt(7))^x is just a small adjustment to (2+sqrt(7))^x. Now to look at the parity of x. If x is odd then (2-sqrt(7))^x is negative and if x is even then (2-sqrt(7))^x is positive.
So then if x is even then floor[(2+sqrt(7))^x] is equal to (2+sqrt(7))^x + (2-sqrt(7))^x - 1, which them means floor[(2+sqrt(7))^x] is odd when x is even.
Similarly, if x is odd then floor[(2+sqrt(7))^x] is equal to (2+sqrt(7))^x + (2-sqrt(7))^x, which them means floor[(2+sqrt(7))^x] is even when x is odd.
Taken together then the sequence floor[(2+sqrt(7))^x] alternates between even and odd integers, which then implies that the sum of two consecutive terms will always be odd.
Solution
