Sorry, an error occurred while loading the content.

## Re: Fermat factorization

Expand Messages
• On 11/11/2012 9:58 AM, kad ... Suppose N = 3 y, where y is prime. Then if a^2 - N = b^2, then a = (y + 3)/2 and b = (y-3)/2. Thus in this worst case, you
Message 1 of 4 , Nov 11, 2012
On 11/11/2012 9:58 AM,

wrote:
> 1. Fermat factorization
> Date: Sun Nov 11, 2012 12:48 am ((PST))
>
> Hi,
> Can any one tell me Whats the exact logic lie behind the worst time complexity of fermat factorization method?
>
>
> In order to factorize N = pq any semiprime in order to do fermat factorization we have to search 'a' > sqrt(N) so that a^2 - N = b^2. Then factorization of N = (a - b)(a + b). My question is when worst running time complexity will occur and whats the reason for worst running time.

Suppose N = 3 y, where y is prime.

Then if a^2 - N = b^2,

then a = (y + 3)/2 and b = (y-3)/2.

Thus in this worst case, you will need to advance "a" to (N/3 + 3)/2
before finding the Fermat factors.

Kermit Rose

[Non-text portions of this message have been removed]
• Hi The absolute limit to which the fermat factorization would have to go to find a solution at the worst would be when the number, N, is prime and no tests
Message 2 of 4 , Nov 12, 2012
Hi
The absolute limit to which the fermat factorization would have
to go to find a solution at the worst would be when the number, N, is
prime and no tests have been carried out: the maximum value would be
given by (N + 1 - (2s))/2, where s= flr(sqrt(N)).

if some trial division has been carried out to say D, the maximum
limit can be reduced to:

(((s - D)^2 + (N - (s*s))/D)/2

The number of operations can be reduced by looking at (N + 1 - (2s))
mod 4, if the result is 0 then only even values need be looked at, and
if the result is 2 then only odd numbers need be looked at.

example N = 13199 s = 114

(13199 + 1 - (2*114)) = 12972

12972 mod 4 = 0

the numbers that need to be looked at are (114 + 2aa) aa =
1, 2, 3, ... upto (12972/2)/2.

example N = 1803601 s = 1342

(1803601 + 1 - (2*1342) = 1800918

1800918 mod 4 = 2

the numbers that need to be looked at are (1342 + (2aa -1)) aa =
1, 2, 3 .... upto (1800918/2)/2

For large numbers the amount of processing would still be
considerable and then one would need to look at a sieve for the values
of a.

Ron

>----Original Message----

>

>wrote:

>> 1. Fermat factorization

>> Date: Sun Nov 11, 2012 12:48 am ((PST))

>>

>> Hi,

>> Can any one tell me Whats the exact logic lie behind the worst time
complexity of fermat factorization method?

>>

>>

>> In order to factorize N = pq any semiprime in order to do fermat
factorization we have to search 'a' > sqrt(N) so that a^2 - N = b^2.
Then factorization of N = (a - b)(a + b). My question is when worst
running time complexity will occur and whats the reason for worst
running time.

>

>Suppose N = 3 y, where y is prime.

>

>Then if a^2 - N = b^2,

>

>then a = (y + 3)/2 and b = (y-3)/2.

>

>Thus in this worst case, you will need to advance "a" to (N/3 + 3)
/2

>before finding the Fermat factors.

>

>

>Kermit Rose
• Dear all He has also sent a message directly to me. He clearly does not understand some very basic concepts: 1) If you think that you have found an error, look
Message 3 of 4 , Nov 25, 2012
Dear all
He has also sent a message directly to me.

He clearly does not understand some very basic concepts:
1) If you think that you have found an error, look for the error
in your own work and if you cannot spot the error put the work away
for sometime (3 months is good) then go back to it.

2) If a message has been posted to a group and no one has found an
error, the likelihood is that the error is in your own work; usually
true 999 times out of a thousand.

3) if the formula work on small numbers, try the same formula on
medium size numbers and than on large numbers; and not just one of
each, it would not be that first time that spurious results can result
from special factors relating to the number(s).

4) Do not use results, ie the difference, which can only be found
after the complete factorization has been completed.

Ron

>----Original Message----

>Date: 23/11/2012 13:24

>To: "ronhallam@..."<ronhallam@...>

>Subj: Re: [PrimeNumbers] Re: Fermat factorization

>

>The logic that factors with least difference will factored easily and
large

>difference will factored hardly is wrong. Please follow the link which
i

>found about Fermat factorization.

>

>On Mon, Nov 12, 2012 at 5:08 PM, ronhallam@... <
>ronhallam@...> wrote:
>
>> **
>>
>>
>> Hi
>> The absolute limit to which the fermat factorization would have
>> to go to find a solution at the worst would be when the number,
N, is

>> prime and no tests have been carried out: the maximum value would
be

>> given by (N + 1 - (2s))/2, where s= flr(sqrt(N)).

>>

>> if some trial division has been carried out to say D, the maximum

>> limit can be reduced to:

>>

>> (((s - D)^2 + (N - (s*s))/D)/2

>>

>> The number of operations can be reduced by looking at (N + 1 -
(2s))

>> mod 4, if the result is 0 then only even values need be looked at,
and

>> if the result is 2 then only odd numbers need be looked at.

>>

>> example N = 13199 s = 114

>>

>> (13199 + 1 - (2*114)) = 12972

>>

>> 12972 mod 4 = 0

>>

>> the numbers that need to be looked at are (114 + 2aa) aa =

>> 1, 2, 3, ... upto (12972/2)/2.

>>

>> example N = 1803601 s = 1342

>>

>> (1803601 + 1 - (2*1342) = 1800918

>>

>> 1800918 mod 4 = 2

>>

>> the numbers that need to be looked at are (1342 + (2aa -1)) aa =

>> 1, 2, 3 .... upto (1800918/2)/2

>>

>> For large numbers the amount of processing would still be

>> considerable and then one would need to look at a sieve for the
values

>> of a.

>>

>> Ron

>>

>>

>> >----Original Message----

>>

>>

>> >

>>

>> >wrote:

>>

>> >> 1. Fermat factorization

>>

>>

>> >> Date: Sun Nov 11, 2012 12:48 am ((PST))

>>

>> >>

>>

>> >> Hi,

>>

>> >> Can any one tell me Whats the exact logic lie behind the worst
time

>> complexity of fermat factorization method?

>>

>> >>

>>

>> >>

>>

>> >> In order to factorize N = pq any semiprime in order to do fermat

>> factorization we have to search 'a' > sqrt(N) so that a^2 - N =
b^2.

>> Then factorization of N = (a - b)(a + b). My question is when worst

>> running time complexity will occur and whats the reason for worst

>> running time.

>>

>> >

>>

>> >Suppose N = 3 y, where y is prime.

>>

>> >

>>

>> >Then if a^2 - N = b^2,

>>

>> >

>>

>> >then a = (y + 3)/2 and b = (y-3)/2.

>>

>> >

>>

>> >Thus in this worst case, you will need to advance "a" to (N/3 + 3)

>> /2

>>

>> >before finding the Fermat factors.

>>

>> >

>>

>> >

>>

>> >Kermit Rose

>>

>>
>>
>
Your message has been successfully submitted and would be delivered to recipients shortly.