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Short factoring proofs: Addition of phi(N)

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#18 2 comments 0 reactions 0 assignees View on GitHub

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Assessment

Difficulty
4/5
Estimated time
3-5 days
Newbie friendliness
25/100
Issue type
Documentation
Clarity
Needs clarification
Activity status
Stale

Research direction

Start at the short factoring proof and the vanilla protocol verifier check described in the issue; verify the exponent algebra and the group-order reasoning. Done when the documentation clearly answers whether replacing N−φ(N) with N+φ(N) is valid and records the corrected explanation if needed.

Written by the indexing model from the issue text.

Description

In the short factoring proof, there is the line $y = r + (N - \varphi(N))e$. In the vanilla protocol, the verifier checks $x \overset{?}{\equiv} z^{y - eN} \mod N$. Because $\varphi(N)$ is the order of the multiplicative group of integers modulo $N$, taking the exponent of a multiple of $\varphi(N)$ gives you the identity 1 for all group elements, so $z^{\varphi(N)e} \equiv 1 \mod N$. This is my understanding how the protocol works.

I was wondering, can we define $y = r + (N + \varphi(N))e$ instead? The minus sign seems redundant to me and the above argument should still work, shouldn't it?

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