- **FermatSearch**
(*https://www.mersenneforum.org/forumdisplay.php?f=133*)

- - **possible overlapping Fermat factor ranges**
(*https://www.mersenneforum.org/showthread.php?t=26338*)

possible overlapping Fermat factor rangesA Fermat number is F(m) = 2^(2^m) + 1.
It has been shown that any factor of a Fermat number has the form k*2^n + 1. with n greater than or equal to m+2 This information is at [URL="fermatsearch.org"]fermatsearch.org[/URL]. if k is not odd then we have an equivalent representation since 2*k*2^c + 1 = k*2^(c+1) + 1 I assume that mmff and other Fermat search programs only search for odd k because the even cases will be searched in increased exponent Also, in the log of known Fermat factors, [URL="http://www.prothsearch.com/fermat.html"]http://www.prothsearch.com/fermat.html[/URL] All the k values are odd. Regards, Matt |

[QUOTE=MattcAnderson;567303]A Fermat number is F(m) = 2^(2^m) + 1.
It has been shown that any factor of a Fermat number has the form k*2^n + 1. with n greater than or equal to m+2 This information is at [URL="fermatsearch.org"]fermatsearch.org[/URL]. if k is not odd then we have an equivalent representation since 2*k*2^c + 1 = k*2^(c+1) + 1 I assume that mmff and other Fermat search programs only search for odd k because the even cases will be searched in increased exponent Also, in the log of known Fermat factors, [URL="http://www.prothsearch.com/fermat.html"]http://www.prothsearch.com/fermat.html[/URL] All the k values are odd. Regards, Matt[/QUOTE] AFAIK, they eliminate even k. I know that gfndsieve does so I assume the others do as well. |

If you think about how modular exponentiation works (square and shift), then you will see why even k's are not needed. To test if some q=k*2^n+1 divides some Fm, you start with 2, and square mod q. If you get -1, then q divides F1. If not, you square again, and if -1, then q divides F2. If not, you square again, and if -1, then q divides F3. If not, you square again, and if -1, then q divides F4. If not, you square again, and if -1, then q divides F5.
Now you see why even k are not relevant? Say I want to see if q=296*2^13+1 divides F11 (n has to be at least m+2 from a well known theorem). Repeating the above square+mod nine times, you get that this q divides F9 and you stop. Because two Fermat numbers can't share a factor. Exactly the same result you will get if you try to test if q=37*2^16+1 divides F14 - you will run into "q divides F9" and stop after 9 steps (this is the same q). |

[QUOTE=LaurV;568454]If you think about how modular exponentiation works (square and shift), then you will see why even k's are not needed. To test if some q=k*2^n+1 divides some Fm, you start with 2, and square mod q. If you get -1, then q divides F1. If not, you square again, and if -1, then q divides F2. If not, you square again, and if -1, then q divides F3. If not, you square again, and if -1, then q divides F4. If not, you square again, and if -1, then q divides F5.
Now you see why even k are not relevant? Say I want to see if q=296*2^13+1 divides F11 (n has to be at least m+2 from a well known theorem). Repeating the above square+mod nine times, you get that this q divides F9 and you stop. Because two Fermat numbers can't share a factor. Exactly the same result you will get if you try to test if q=37*2^16+1 divides F14 - you will run into "q divides F9" and stop after 9 steps (this is the same q).[/QUOTE] To elaborate a bit more, any k*2^n+1 with even k can always be represented as k'*2^n'+1 with k' odd and n' > n. For example, LaurV's 296*2^13+1 = 37*2^16+1. Most Fermat factor search programs will test a range of k for a particular value of n, then test the same range of k for n+1, etc. So testing even values of k for n is unnecessary because the corresponding k*2^n+1 will eventually be tested by you (or some other researcher) using an odd k with a larger n. |

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