Deep dive into rsa

Astory
Genesys confidential and proprietary information.
Unauthorized disclosure is prohibited. 2
Bjarne Stroustrup Linus Torvalds
Dennis Ritchie

Ropeproblem
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Unauthorized disclosure is prohibited.
Extend both ropes by 1 foot

Ropeproblem
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C1 = 2π R1
C2 = 2π R2
(R2-R1) = (C2-C1)/2π
= 1 foot/2π

Ropeproblem
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RSA– Keygeneration
◦ Choose two primes p and q of approximately equal size so that their product n = pq is
of the required length (1024, 2048…).
◦ Compute φ(n) = (p-1)(q-1).
◦ Choose a public exponent e, 1 < e < φ(n), which is coprime to φ(n), that is, gcd(e,
φ(n))=1. coprime {4, 9}, {8, 21}
◦ Compute a private exponent d that satisfies the congruence ed ≡ 1 (mod φ(n)).
ed = 1+φ(n)*k
◦ Make the public key (n, e) available to others. Keep the private values d, p, q, and
φ(n) secret.
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RSA– Encryption
◦ Encryption c = me mod n, 1 < m < n-1
◦ Decryption m = cd mod n
◦ m = RSAPUBLIC(RSAPRIVATE(m))
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RSA– Period
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p=3, q=5, n=15, φ(n)=(p-1)(q-1)=8, m=2
n 1 2 3 4 5 6 7 8 9 10 11 12
2n 2 4 8 16 32 64 128 256 512 1024 2048 4096
2n mod 15 2 4 8 2 4 8 2 4 8
p=3, q=7, n=21, φ(n)=(p-1)(q-1)=12, m=2
n 1 2 3 4 5 6 7 8 9 10 11 12
2n 2 4 8 16 32 64 128 256 512 1024 2048 4096
2n mod 21 2 4 8 16 11 2 4 8 16 11
φ
φλ
λ

RSA -φ(n)
◦ φ(n) - how many positive integers less than n and coprime to n.
E.g. φ(12)=4 as the 4 integers {1,5,7,11}
◦ For any prime p, φ(p)=p−1, since all numbers less than p are coprime to it.
E.g. φ(7)=6 as the 6 integers {1,2,3,4,5,6}
◦ If m and n are coprime, then φ(m)φ(n)=φ(mn).
Now we understand what is ‘Compute φ(n) = (p-1)(q-1)’
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RSA- Pattern
p=7, φ(p)=6, m=2
i 1 2 3 4 5 6
i*m 2 4 6 8 10 12
i*m mod p 2 4 6 3 5
p=12, φ(p)=4, m=5 (m coprime to p)
i 1 5 7 11
i*m 5 25 35 55
i*m mod p 5 1 11

RSA- Fermat’sLittleTheorem
p=7, φ(p)=6, m=2
i 1 2 3 4 5 6
i*m 2 4 6 8 10 12
i*m mod p 2 4 6 3 5
m*2m*3m…*(p-1)m ≡ 1*2*3…*(p-1)(mod p) =>
(mp-1)*1*2*3…*(p-1) ≡ 1*2*3…*(p-1)(mod p) =>
mp-1 ≡ 1 (mod p) => mφ(n) ≡ 1 (mod p)

RSA -Proof
◦ Encryption c = me mod n
◦ Decryption m = cd mod n
◦ ed = 1+kφ(n)
◦ cd ≡ med
≡ m1+kφ(n)
≡ m.(mφ(n))k
≡ m.1k ,since mφ(n) ≡ 1 (mod n)
≡ m (mod n).
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RSA -Breaking
◦ me mod n with small e, m is not secure (if me < n)
◦ Today almost in all public keys (e,n), e=3, 17 and 65537, why?
c1 = m3 mod n1, c2 = m3 mod n2, c3 = m3 mod n3
m can be revealed by Chinese remainder theorem
c1* c2* c3 = m3 mod (n1* n2* n3) = m3 (if m3 < n1* n2* n3 )
◦ RSA without padding is vulnerable
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RSA -Breaking
◦ Brute force - Check all odd numbers starting below the square root of n.
◦ Euclidean Algorithm – if you can find two n1 and n2 have a common factor. In the real
world, about two out of 1000 secure web sites RSA moduli do have common factors,
and thus are susceptible to cracking (http://eprint.iacr.org/2012/064.pdf)
◦ Shor's algorithm -Key is to find ‘Period’ which requires a quantum computer
◦ https://en.wikipedia.org/wiki/RSA_Factoring_Challenge
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RSA -Quantumsuperposition
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RSA -Quantumentanglement
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RSA -Quantumcomputer
◦ Qubits – Superposition and Entanglement (n qubits can represent 2n state)
◦ Qubits – State is described in probability amplitude (0.5, 0.866…), (0.5, -0.866…)
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Bloch sphere

RSA -Shor'salgorithm
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Quantum Fourier Transform analogy

Questions?
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