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tl;dr version of the 2016 ACM Turing Award: ((g^a) mod p)^b mod p = (g^ab) mod p = ((g^b) mod p)^a mod p -- The beauty of math.
by wbsun 11y ago
tl;dr version of the 2016 ACM Turing Award:
((g^a) mod p)^b mod p = (g^ab) mod p = ((g^b) mod p)^a mod p
-- The beauty of math.
- AstralStorm 11y agoThis skips the most important part of the proof which explains why this is not easily invertible without the secret key.
- lkowalcz 11y agoWhat is this part? Isn't it just by assumption? (also, by secret key, I'm guessing you mean 'a', 'b')