Show that for any integer value of n > 0:

for some positive integer N.

Let:

( + 1)^{k} = a_{k} + b_{k}

Note that:

(a_{k} + b_{k})( + 1)

= 2a_{k} + a_{k} + b_{k} + b_{k}

= (a_{k} + b_{k}) + (2a_{k} + b_{k})

Hence:

a_{k+1} = a_{k} + b_{k}

b_{k+1} = 2a_{k} + b_{k}

Now let's show that:

2a_{k}^{2} - b_{k}^{2} = 1 or -1

We'll prove it by induction. For the case n = 1:

a_{1} = 1

b_{1} = 1

2a_{k}^{2} - b_{k}^{2} = 2 - 1 = 1

Assume for n = k:

2a_{k}^{2} - b_{k}^{2} = 1 or -1

Now for n = k+1:

2a_{k+1}^{2} - b_{k+1}^{2}

= 2(a_{k} + b_{k})^{2} - (2a_{k} + b_{k})^{2}

= 2a_{k}^{2} + 4a_{k}b_{k} + 2b_{k}^{2} - 4a_{k}^{2} - 4a_{k}b_{k} - b_{k}^{2}

= b_{k}^{2} - 2a_{k}^{2}

= -(2a_{k}^{2} - b_{k}^{2})

= 1 or -1, by the inductive hypothesis

Now let N be the smaller of 2a_{k}^{2} and b_{k}^{2}.
The other value is N+1, by the result just proven above.
Then:

( + 1)^{k}

= a_{k} + b_{k}

=