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1  Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it on: May 10, 2024, 08:19:01 PM
hi all i read the post since one year sorry for my english i found a interesting think but i does takes me further. in every elliptic curve like y^2= x^3+7 there is something interesting like :
if P(1,y1) - k times--> Q(-29/3 ,y2)
   P(2,y3) --k times--> Q(-3,y2)
 so on there is a simple math here where k is always too know independent from whcih curve we work .
 i don't want to give more information this operation is 10 times faster then k*G= and find the x value 
2  Bitcoin / Development & Technical Discussion / Re: Pollard's kangaroo ECDLP solver on: September 22, 2023, 09:56:06 AM
is there a way for search y coordinate square modulo p
3  Bitcoin / Bitcoin Discussion / Re: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== on: June 30, 2023, 09:30:51 AM
Why there is no activity around these woods? Chop chop people, get to work.
If it helps, I am 99.99% sure that puzzle #125 starts with 1a. I just don't know how could that be of any help other than lowering the bit range, but if there is any secret behind it, do let me know.😉

Something to work on. Please do your calculations and tell me if I'm wrong.

2000 (2^125) = 037e2cd40ef8c94077f44b1d1548425e3d7e125be646707bad2818b0eda7dc0151
1700 = (puzzle #125) = 0233709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e
300 = 0286936a275e6d53bb2b2718c93d8a5aa44f371f6e0300abb73b89dd851d2fbe88
700 = 03ed01ff219ed5c1afc12d991a82e3063ddcee1fd53b46f7cad52a0d87a7112aed
400 = 02a64a0b3739ddccddece6d90407c925717c75467cc8ce46321d73ec2663320130
200 = 0339ddd9a2a1a113c105175e17903c1f72326ff89b109efc8b976cc9916429c9c4
100 = 031f45d50a743e772f27543272ff4aba36da659540af3185907ad08e68ed0eee4f

And here is 500 = 2^123

500 = 020bfc0504a4b3235d065c0d426b8675fcb2c85d6f58275d791b43e1fe44a6db03

0000000000000000000000000000000008000000000000000000000000000000


I will leave a quote from gmaxwell here, I don't know whether my calculations are proving the #125 is in the range between 2^124 and 2^125 or not?

With the help of someone knowing the secret they could prove it was in a range using a confidential-transactions like zero knowledge range proof. (which is exactly what CT does... proves the values are in  a range like [0,2^32) that couldn't overflow.)






what calculation do you have made
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