Here's an informal proof of our conjecture "The factor of N closest to sqrt(N) is less than sqrt(N)".
 Let M be the positive square root of N. (M*M = N, M>0)
 The closest factor to M is bewteen 1 and 2*M. (i.e. 1 is closer to M than any number greater than twice M. (M1 < 2*MM))
 For every pair of numbers whose product is N, one of the pair will be greater than M and the other less than M. (the product of two numbers greater than M is greater than N, and the product of two numbers less than M would be less than N)
 Take a pair of numbers whose product is N. Call the smaller (MX) and the larger (M+Y). (i.e. X>0, Y>0)
 So, (MX) * (M+Y) = N
 Multiplying out, M^2 + M*(YX)  X*Y = N
 But M^2 = N (by definition)
 Substituting: N + M*(YX)  X*Y = N
 Subtracting N: M*(YX)  X*Y = 0
 Since X and Y are positive (by definition, step 4), the term X*Y is negative
 If X*Y is negative, the only way for the entire sum to equal 0 is if the term M*(YX) is positive
 But M is positive (step 1), so (YX) must also be positive, which means that Y is greater than X.
 Finally, if Y is greater than X, the smaller factor, (MX) is closer to M than (M+Y), so you can limit your factor search to numbers less than sqrt(N).
 All code is 100% tested and functional unless otherwise noted.
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