QUANTUM 08
How finding a hidden period breaks RSA
Sinclair College - Dr. Mike Libassi
TODAY
01 · RSA REFRESHER
Two secret primes. Example: 3 and 5.
Public modulus. Example: 15.
(p − 1)(q − 1). Example: 8.
Example: e = 3, d = 11, since 33 mod 8 = 1.
02 · PERIOD FINDING
Random number, e.g. a = 7
7 mod 15 = 7, 49 mod 15 = 4
343 mod 15 = 13, 2401 mod 15 = 1
Back to 1 after r = 4 steps
Listing terms one by one is the slow classical way. The quantum computer finds r without listing them.
03 · THE QUANTUM STEP
Interference concentrates the probability on outcomes spaced Q/r apart. That's the same trick as Deutsch-Jozsa, scaled up.
04 · FROM PERIOD TO FACTORS
r = 4 for a = 7
7² mod 15 = 4
gcd(3, 15) = 3 and gcd(5, 15) = 5
15 = 3 × 5
Not every a works. If r is odd, or ar/2 ≡ −1 (mod N), pick a different a and run again. Most choices succeed within a few tries.
04 · REAL-WORLD SCALE
| Estimate | Qubits needed | Run time |
|---|---|---|
| Gidney & Ekerå, 2019 | ≈ 20 million noisy physical | ≈ 8 hours |
| Gidney, 2025 | under 1 million noisy physical | under a week |
| Today's processors | on the order of 1,000 physical | no large-scale error correction yet |
The estimate dropped about twentyfold in six years through better algorithms, not bigger hardware.
WRAP-UP
Factoring N exposes the RSA private key d
Shor's algorithm reduces factoring to finding the period r of aˣ mod N
Superposition plus the QFT reveals r efficiently
Resource estimates keep falling, so the migration to PQC can't wait
DISCUSSION
Estimates for breaking RSA dropped twentyfold in six years without new hardware. What does that suggest about planning timelines?
Connect this to "harvest now, decrypt later" in Quantum 09.