The golden ratio number φ = (1+√5)/2 possesses many interesting properties.

inter alia
For any even integer n: φⁿ + 1 /φⁿ is an integer
For any odd integer n: φⁿ - 1 /φⁿ is an integer

Prove the above.

Let x= 1/φ, so that x^2+x = 1.

Small values of the above expressions are readily computed and shown to take integer values, e.g.

1 (1/φ+1)^(1)-(1/φ)^(1) = 1  
2 (1/φ+1)^(3)-(1/φ)^(3) = 4  
3 (1/φ+1)^(5)-(1/φ)^(5) = 11  
4 (1/φ+1)^(7)-(1/φ)^(7) = 29  
 etc.  A002878
 
 (1/φ+1)^(0)+(1/φ)^(0) = 2  
5 (1/φ+1)^(2)+(1/φ)^(2) = 3  
6 (1/φ+1)^(4)+(1/φ)^(4) = 7  
7 (1/φ+1)^(6)+(1/φ)^(6) = 18  
8 (1/φ+1)^(8)+(1/φ)^(8) = 47
 etc. A005248

We are going to deduct 2-5 = 1, 3-6 = 2 , 4-7 = 3, etc. These are arithmetic operations and so commutative.

Quite generally:  (x+1)^n-x^n-((x+1)^(n-1)+x^(n-1)) = x (x+1)^(n-1)-(x+1) x^(n-1); when  x= 1/φ, we can simplify this to  (x+1)^(n-2)-x^(n-2) by dividing each part by x(x+1) which we know is 1, thereby justifying the integer recurrences if the first few values are known.

Contrariwise, if the initial values are, say,

1 (1/φ+1)^(1)+(1/φ)^(1) = sqrt(5)    
2 (1/φ+1)^(3)+(1/φ)^(3) =2sqrt(5)    
3 (1/φ+1)^(5)+(1/φ)^(5) = 5sqrt(5)    
4 (1/φ+1)^(7)+(1/φ)^(7) = 13sqrt(5)    
 etc.  A001519
   
 (1/φ+1)^(0)-(1/φ)^(0) = 0    
5 (1/φ+1)^(2)-(1/φ)^(2) = sqrt(5)    
6 (1/φ+1)^(4)-(1/φ)^(4) =3sqrt(5)    
7 (1/φ+1)^(6)-(1/φ)^(6) =8sqrt(5)    
8 (1/φ+1)^(8)-(1/φ)^(8) =21sqrt(5) 
 etc. A001906

then integer solutions can never occur in the sequence. The explanation is that the expansion of (x+1)^n-x^n-((x+1)^(n-1)+x^(n-1)) always contains the factor x(x+1) = 1 in the first case, while the corresponding factor in the second is (2x+1), or sqrt(5).

Posted by broll on 2016-10-18 05:25:42