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Comment by wmf | Hacker News Reader

Primary source if anyone's interested: https://cyber.stanford.edu/sites/default/files/bramcohen.pdf https://www.youtube.com/watch?v=aYG0NxoG7yw&feature=youtu.be

Of course, the Internet doesn’t remember that he worked on cryptocurrency almost 20 years ago (with Zooko of ZCash) before he started BitTorrent. Pepperidge Farm remembers.

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Actually the primary source - at least for this article - is the latest episode of STEAL THIS SHOW: https://stealthisshow.com/s02e14. The interview has some detail on when he'll start the enterprise et cetera.

We also go into the pre-history of BitTorrent at Mojo Nation.

Good show. In response to something Bram mentions offhand I do have to retort that when Bram suggested changing the MN architecture to be more like what was eventually to become BitTorrent the response, which in hindsight I will admit was short-sighted and incredibly dumb, was due to having too little runway left to completely rewrite an already over-engineered codebase and not any sort of crypto-libertarian ideals. (And yes, I keep a copy of his email message suggesting this change to our internal list posted above my desk as a constant reminder...)

Does anyone have more details on slide 9 "Proof of time: Modular roots"?

> In group of order 2^k, square k-1 times to find a square root

I don't want to say anything stupid, but I don't know if this is trivially true, for instance in [the multiplicative group of] a field of char 2, square is linear and so it's square root.

Edit: to clarify, in a field of char 2 the multiplicative group has order 2^q-1, which can't be 2^k. I'm just giving this as an example of a group where square root is fast.

I also fine this confusing.

From the video, it sounds like the claim is that sqrt(x) = x^(2^(k-1)), which you verify by checking that x=x^(2^k) [0]. However, since the order of the group is 2^k, it is an elementary result of group theory that x^(2^k)=1 [1]

[0] That is, you square the claimed root which should be x^(2^(k-1)).

EDIT: watching the video some more for clues, he does mention that he uses a particular kind of group, so the claimed property does not need to hold for any group of order 2^k. However, as I argue above, I still don't see how it is possible for any group of order 2^k.

[1] This is a direct result of Lagranges theorem. https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_th...

getting the square root of x from k-1 squarings means (x^(2^(k-1))) ^ 2 = x, or x^(2^k) = x. so x has multiplicative order (2^k)-1, which is the order of GF(2^k) under multiplication.

But the group has order 2^k. Since the powers of x (for any group element x) form a group themselves, Lagrange's theorme tells us that the order of x must divide 2^k. This should mean that it is impossible for x to have order (2^k)-1. If fact, it means that x^(2^k)=1.

Further, if you are given the value y=x^2, then no power of y will give you x.

The finite field GF(2^k) has order (number of elements) 2^k, but its multiplicative group has the (2^k)-1 nonzero elements, and thus order (2^k)-1, which squares [pun intended] with the fact that x^(2^k-1) = 1.

"Of course, the Internet doesn’t remember that he worked on cryptocurrency almost 20 years ago (with Zooko of ZCash) before he started BitTorrent."

Citation needed.

Try MojoNation list archives.

Yeah, and Mojonation with EGfaBT before that... although not really with POW. https://en.wikipedia.org/wiki/Mnet_(peer-to-peer_network)

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