in reply to
Re^3: OT: Finding Factor Closest To Square Root

in thread OT: Finding Factor Closest To Square Root

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.
`(M-1 < 2*M-M)`)
- 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 (
`M-X`) and the larger (`M+Y`). (i.e. `X>0, Y>0`)
- So,
`(M-X) * (M+Y) = N`
- Multiplying out,
`M^2 + M*(Y-X) - X*Y = N`
- But
`M^2 = N` (by definition)
- Substituting:
`N + M*(Y-X) - X*Y = N `
- Subtracting N:
`M*(Y-X) - 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*(Y-X)` is positive
- But M is positive (step 1), so
`(Y-X)` must also be positive, which means that Y is greater than X.
- Finally, if Y is greater than X, the smaller factor,
`(M-X)` is closer to M than `(M+Y)`, so you can limit your factor search to numbers less than sqrt(N).

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