RSA Hastad's broadcast attack with large numbers

SJ19

8/17/23, 1:49 PM

I understand the theory behind Hastad's broadcast attack.

Namely if we have three encrypted messages with the exponent e=3:

c1 = m1 mod n1,
c2 = m2 mod n2,
c3 = m3 mod n3

Then we can use the Chinese Remainder Theorem to find

c = c1 mod n1,
c = c2 mod n2,
c = c3 mod n3,
c = m^3 mod n1*n2*n3

and since n1 * n2 * n3 is too large, we simply have c = m^3

I only find explanations for smaller numbers but how are you supposed to find the solution for very large numbers such as

16833444999714344947074933154092703072048227929941882928373643621000348494347

0 + 0

rsa

attack

visual-cryptography

poncho

8/17/23, 4:20 PM

Are you asking about how to compute a cube root (over the integers)?

fgrieu

8/18/23, 6:00 PM

Python works fine with large numbers. Should everything else be unobtainium, good old dichotomy will compute cube root

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: RSA Hastad's broadcast attack with large numbers

TH: การโจมตีออกอากาศของ RSA Hastad ด้วยตัวเลขจำนวนมาก

RO: Atacul difuzat de RSA Hastad cu număr mare

RU: Широковещательная атака RSA Hastad с большими числами

VI: Cuộc tấn công phát sóng của RSA Hastad với số lượng lớn

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