The Basic Method
One tries various values of a, hoping that is a square.
FermatFactor(N): // N should be odd a ← ceil(sqrt(N)) b2 ← a*a - N while b2 isn't a square: a ← a + 1 // equivalently: b2 ← b2 + 2*a + 1 b2 ← a*a - N // a ← a + 1 endwhile return a - sqrt(b2) // or a + sqrt(b2)For example, to factor, one computes
| a: | 78 | 79 | 80 |
|---|---|---|---|
| b2: | 125 | 282 | 441 |
The third try produces a square., and the factors are, and .
Suppose N has more than two prime factors. That procedure first finds the factorization with the least values of a and b. That is, is the smallest factor ≥ the square-root of N. And so is the largest factor ≤ root-N. If the procedure finds, that shows that N is prime.
For, let c be the largest subroot factor., so the number of steps is approximately .
If N is prime (so that ), one needs steps! This is a bad way to prove primality. But if N has a factor close to its square-root, the method works quickly. More precisely, if c differs less than from the method requires only one step. Note, that this is independent of the size of N.
Read more about this topic: Fermat's Factorization Method
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