Re: Factoring problem solution

anonymous_at_mathforum.org
Date: 02/11/05 

Date: Fri, 11 Feb 2005 20:44:25 +0000 (UTC)

On 11 Feb 2005, Jack Fearnley wrote:
>Johann Wiesenbauer wrote:
>
>>>x^2 - y^2 == 0 (mod N) and x != y then you have a good chance of
>>>factoring N.
>>>
>>>Tom
>>
>> When demanding x != +/-y (mod N) this chance becomes even 100%.
>>
>> Johann
>
>Not if x-y=1 and x+y=N
>
>Jack

If x+y=N, then x = -y (mod N). Hence, this case is included in my
condition.

Johann