try this...

For every r > 1 , I assume that ,

r + 1r = a

Again , [ r ] = r + 1r - 1

So , [ r ^m ] = r ^m + 1r ^m - 1

= a _m - 1

So , [ r ^m ] + 1 = a _m - 1 + 1 = a _m ≡ 0 ( mod k ) ...........Given

Now , a ₀ = 2 ; a ₁ = a ; a _{n + 2} = a a _{n + 1} - a _n

Obviously , we can choose such an " a _m " which satisfies the above relations .

But for that to happen , we must have ,

a ≡ 0 ( mod k )

So , lets put a = n k .

Again , we can choose exactly such an " a " for which D = a ² - 4 is not a perfect square also .

Hence , we get our required " r " in terms of k .

r = a + √D2 = 12 { n k + ( n ² k ² - 4 ) ^{1 / 2} }