2025/009) Show that $(2+\sqrt{5} )^n + (2-\sqrt{5} )^n$ is integer for $n\in Z^+$

 We shall prove the same by induction

For n = 1 we have $(2+\sqrt{5} )^1 + (2-\sqrt{5} )^1 = (2+\sqrt{5} ) + (2-\sqrt{5} ) = 4 $

For n = 2 we have $(2+\sqrt{5} )^2 + (2-\sqrt{5} )^2 = (4+4\sqrt{5} + 5) + (4-4\sqrt{5} +5) = 18$

Let us assume that it is true for k upto n. We shall show that it is  true for k = n+1

below we shall keep power 1 for clarity.

Let $T_n = (2+\sqrt{5} )^n + (2-\sqrt{5} )^n$

We have $T_nT_1 = ((2+\sqrt{5} )^n + (2-\sqrt{5} )^n) ((2+\sqrt{5} )^1 + (2-\sqrt{5} )^1)$

$= (2+\sqrt{5} )^{n+1} + (2+\sqrt{5} )^n * (2-\sqrt{5} )^1 + (2-\sqrt{5} )^1(2-\sqrt{5} )^n + (2-\sqrt{5} )^{n+1}$

$= ((2+\sqrt{5} )^{n+1} +  (2-\sqrt{5} )^{n+1})+ (2+\sqrt{5} )^n * (2-\sqrt{5} )^1 + (2-\sqrt{5} )^1(2-\sqrt{5} )^n $

$=T_{n+1}+ (2+\sqrt{5} )^n * (2-\sqrt{5} )^1 + (2-\sqrt{5} )^1(2-\sqrt{5} )^n $ using definition

$=T_{n+1}+ (2+\sqrt{5} )^{n-1} *  (2+\sqrt{5} ) * (2-\sqrt{5} )^1 + (2-\sqrt{5} )^1 (2-\sqrt{5} )(2-\sqrt{5} )^{n-1} $ 

$=T_{n+1}+ (2+\sqrt{5} )^{n-1} *  (-1) -1 * (2-\sqrt{5} )^{n-1} $ using product

$=T_{n+1} -1 ((2+\sqrt{5} )^{n-1}+ (2-\sqrt{5} )^{n-1}$ using T

$=T_{n+1} - T_{n-1}$

Or $T_{n+1} = T_nT_1 + T_{n-1}$

or $T_{n+1} = 4 T_n + T_{n-1}$ using the value of $T_1$'

so if it is integer for value upto n then it is integer for k = n+1

Proved