Can anyone solve this that i call paradox?
Most of you are familiar with the formula:
s1 x k1 = z1 + r1 x d1 mod n
If you have two of them related like this:
s1 x k1 = z1 + r1 x d1 mod n (1)
s2 x k1 = z2 + r1 x d1 mod n (2)
There is the formula:
(s1 - s2) x k1 = (z1 - z2) mod n (3)
to find,k1.And from that d1.
There also special cases then you can find d1,from the formula:
k2 = k1 - 1 or m,m < n (or)
k2 = k1 /2 or m,m < n.
Also,i read in many posts here,that k1 and k2 or (in special cases) d1 and d2,can not be related in a algebric and logic way,beacouse of course,will be a formula that will find k1,k2,d1 and d2. Now,i found with a simple derivation a formula:
s1 x k1 = z1 + r1 x d1 (4)
s2 x k2 = z2 + r2 x d1 (5)
a x k1 - b x k2 = c (6)
Now the paradox,from my point of view:
I admit that my advanced algebra is weak,but i want to make sure that if someone is better then me can help me to prove i am right or prove me wrong and help me and in the end help you,the reader.
I can prove without,that i call invisible info/stealth info,that this formula is true.By invisible info /stealth info,i mean to have two real or artificial tx (transactions),where i can use invisible info aka real value of k1,k2 or d1,to check the information directly.
But here is the thing: even i can prove in any transactions anywhere,relation between a x k1 and b x k2 is c,by means of numbers,logic and formula(4),(5) and (6) only,without invisible info,i can not extract k1,k2 or d,even in a two related k1,k2 transaction made public here in this forum.
I tried two modes to extract k1,k2 or d,but it gave me zero division. Why,if i can prove with algebra and logic that are related?
I think first because my advanced algebra is not high enough or second can be a rule that can not be broken related with how those equations are made.Can you help me?
Note.I am aware that this kind of info is highly sensitive.For that i not made public how i can prove without invisible info,how can k1 and k2 are related by means of (4),(5) and (6).But as you expect with a bit of thinking you can do,too.If someone can give me the answer in private,please do.
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