Cryptography - coin flipping protocol

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Homework Help Overview

The discussion revolves around a cryptographic coin flipping protocol involving modular arithmetic and properties of quadratic residues. Participants explore the implications of certain conditions on primes and their relationships in the context of the protocol.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants examine the conditions under which certain equations hold true, particularly regarding the square roots of y modulo p and q. Questions arise about the assumptions of equality between variables in different modular contexts.

Discussion Status

There is ongoing exploration of the implications of the conditions set forth in the problem. Some participants have provided updates on their reasoning, particularly regarding the impossibility of y being a square modulo both p and q. Others express uncertainty about how to effectively utilize the information given in the problem.

Contextual Notes

Participants note the specific modular conditions of the primes and the implications for the square roots involved in the protocol. There is mention of the need for factorization of n and the consequences of Bob's ability to produce p and q.

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Homework Statement
coin flipping protocol
Relevant Equations
n = pq
b^2 = y(modp)
b^2 = -y(modq)
Screen Shot 2020-11-02 at 2.37.00 AM.png

Conditions of the problem:
y =/= a^2 (mod pq)
-y =/= a^2 (mod pq)
p = 3 (mod 4)
q = 3 (mod 4)

(a) We assume y is both square (mod p) and (mod q). Then,
b^2 = y (mod p)
b^2 = y (mod q)
b^2 = y + pk
b^2 = y + qr
pk = qr ?

(should we assume b is the same between the two equations?)

(b)
-y = d^2 (modq)
q-y = d^2 (modq)
y + d^2 = q (modq)
d^2 = -y (modq)

(c)
y = a^2 (mod p)
-y = a^2 (mod q)

(again, should we assume b is the same between two equations?)

(d)
If bob has b and y, he can compute a relationship between p and q.
b^2 = y + pk
b^2 = -y + qr
2 b^2 + pk + qr ?

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(description of the coin flipping protocol)

Alice knows primes p = 3(mod) and q = 3(mod4). She computes pq = n, she sends n to Bob.

Bob chooses u so y = u^2 (modn), sends y to Alice

Alice finds 4 square roots of y using p and q.
square roots = u, -u, v, -v,

Alice picks one square root at random and sends it to Bob. If Alice picks u or -u she wins. If she picks v or -v, she loses.

Bob cannot lie and falsify Alice's loss because he only knows u, -u. therefore he will not be able to provide v if Alice asks for it.
 
Last edited:
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Edit: At the end of the protocol, Alice asks Bob for the factorization of n. If Bob can produce p and q, it means he has all four square roots and Alice lost. If Bob can't produce p and q, it means Alice won.
 
update on (a)

if y is square mod p and q, we have the system of equations

x^2 = y (mod p)
x^2 = y (mod q)

which is only true if

x^2 = y + p*q
x^2 = y + q*p

however,

x^2 =/= y (mod n)

and

x^2 =/= y + pq*r for some constant r. in this case, we have r = 1

so y can't be a square both mod p and mod q, We use the same procedure to show for -y.
 
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(b)

x^2 = -y (mod q)
x^2 = q-y (mod q)
x^2 + y = 0 (mod q)
x^2 = -y (mod q)
This seems like I'm going around in circles, it feels like i haven't used the relevant information. Does anyone see what i should be doing?

(c)
I'm not sure how to incorporate the right information to prove what the problem is aking. this sounds extremely general to me. any pointers?

(d). could he use b^2 as one of the square roots (mod n) and find gcd(x-b, n) to factor n?
 

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