Clearly, for any given S_(i+1)(x) to correspond to an integer, it is necessary that S_(i)(x) is an integer.
So, each of S_(i)(x) for i = 1,2,....,2007 must be integers.
But S_2(x)
= sqrt(x+(S_1(x))) ;
= sqrt(S_1(x) + S_1(x)^2); and so:
S_2(x)^2 = S_1(x)(S_1(x) +1)).
Since both S_1(x) and S_2(x) must be integers, this is possible only if:
x = S_1(x) = S_2(x) = 0.
This gives S_(i)(x) = 0 for all i = 1,...,2007 and y=0
Consequently, (x,y) = (0,0) constitutes the only possible solution to Part A.
PART (B):
We observe that:
S_1(x) = sqrt(2x);
S_2(x) = sqrt(x+sqrt(2x)) = sqrt(x+(S_1(x))) ;
S_3(x) = sqrt( x+ (S_2(x))
S_(i+1)(x) = sqrt( x+ (S_i(x))); for i = 2,3,�, 2006.
Clearly, for any given S_(i+1)(x) to correspond to an integer, it is necessary that S_(i)(x) is an integer.
Now, S_1(x) is an integer only when x =2* k^2, for some integer k.
Accordingly, S_2(x) = sqrt(2k(k+1)).
Clearly, for y to correspond to an integer, it follows that S_2(p) must be an integer; so that:
2k(k+1) = (S_2(x))^2
Solving the underlying Pell�s equation for k<300; we obtain:
S_1(x) = S_i(x) = 2, whenever p=2; for i = 2,3,�,2007
Since S_3(x) does not correspond to an integer for x< 2*300^2 = 180,000; we can safely conclude that the equation (#) admits of the solution:
(x,y) = (0,0); (2,2) whenever x<180,000.
