Quote from: bytcoin on February 16, 2020, 12:15:19 AM

One question ... how did you discover the value of a and b?

Simply you posted that values and I have seen them before you obfuscated.

Quote from: bytcoin on February 16, 2020, 12:15:19 AM

Discover this!

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = Huh

??
b =

??

w = GF(p)

w (((a / b)))
w (((b / a)))

w= 41689968412277401815192940146121615515073176981576112778517871276338175969527
w= 94599322545067982882578206403521588574602656593708975301033069273870448067861

I will send you 1 BTC if you find the value a and b

If you mean: retrieve 'a' and 'b' you used from 'w1' and 'w2' is not possible, because given w1 and w2 there is no only a couple (a,b) but many many couples with a/b = w1 and b/a = w2 mod p

For example, for each value of a:

b = w2 * a mod p because b = (b/a) * a mod p

Code:

a = 1 b = 94599322545067982882578206403521588574602656593708975301033069273870448067861
a = 2 b = 73406555852819770341585427798355269296367748908343046219460975406222734641385
a = 10 b = 19656511552150265437214183965712622923326051704490517949489387606559188723914
a = 50 b = 98282557760751327186070919828563114616630258522452589747446938032795943619570
a = 99999999999999 b = 56876268607704945280095310365193269978753984617363817957513687854617877809261
a = 10000000000000000000000000000000000000000000000000000000000000000000000000000 b = 10939142339967927822460418038112020591422658251099346849357313458001062196732

For all these couples (a,b) you have that a/b = w1 mod p and b/a = w2 mod p.

Besides if you provide w1, you don't need to provide w2 too, because w2 is (w1)^-1 mod p --> w2 = 1/w1

w1 = 41689968412277401815192940146121615515073176981576112778517871276338175969527
w2 = pow(w1, p-2, p)
w2
94599322545067982882578206403521588574602656593708975301033069273870448067861

You have to provide w1 and a or b , but from w1 and w2 you cannot get a specific 'a' and 'b'.

In other terms you are providing only a single information (w1) and you want to know 2 values, it is like to resolve a single equation with 2 variables, like y = 2 * x (in your case b = w2 * a mod p)

there is no a single solution, but infinite couples (1,2), (2,4), (3,6) ..... there is no way to guess the couple you used!