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State Fermat algorithm

The basic Fermat algorithm is as follows:

Assume that n is an odd positive integer. Set c = [√n] (`ceiling of √n '). Then we consider in turn the numbers

c2 - n; (c+1)2 - n; (c+2)2 - n.....

until a perfect square is found. If this occurs at the term (c+k)2 - n, then putting a = c+k we have

a2 - n = (c+k)2 - n = b2;

say, and then n = a2 - b2, as desired.

This process will terminate in the worst case when a = (n+1)/2, since

n = [(n+1)/2]2 - [(n-1)/2]2
 
In particular, the number of steps taken will be at worst

(n+1)/2 - [√n] +1 = O(n)

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