This problem is asking for the minimum x such that (x,y) is a solution to the Pells Equation x^2 - Dy^2 = 1, and luckily on the same page we can find a way to solve the equation using continued fractions, see Problem 64 for an explanation on this.
The following theorem is stated in the above linked pages
We can start with h(1) = a(0) because we know that we are not dealing with any cases where sqrt(N) is a perfect square.
Then for each D that is not a perfect square, we find the first solution using h(n) and k(n) and add [x, D] to a list, which we can then sort
Enter an integer (yourinput)
Code will output [x, D] where x is the largest x such that x^2 - Dy^2 = 1 is minimal for all D <= yourinput