How is Chinese remainder theorem used in RSA?

Chinese Remainder Theorem in RSA-CRT In RSA-CRT, it is a common practice to employ the Chinese Remainder Theo- rem during decryption. It results in a decryption much faster than modular exponen- tiation. RSA-CRT differs from the standard RSA in key generation and decryption.

How do you use the Chinese Remainder Theorem examples?

Example: Solve the simultaneous congruences x ≡ 6 (mod 11), x ≡ 13 (mod 16), x ≡ 9 (mod 21), x ≡ 19 (mod 25). Solution: Since 11, 16, 21, and 25 are pairwise relatively prime, the Chinese Remainder Theorem tells us that there is a unique solution modulo m, where m = 11⋅16⋅21⋅25 = 92400.

What is CRT in RSA?

The RSA-CRT domain is composed of an RSA public key (N,e) and an RSA private key (p, q, dp,dq,iq) where N = pq, p and q are large prime integers, gcd((p−1),e) = gcd((q−1),e) = 1, dp = e−1 mod (p − 1), dq = e−1 mod (q − 1) and iq = q−1 mod p.

What is CRT encryption?

This scheme is the one that uses little or limited computation power for encryption and much lesser requirements for decryption. …

How to write out the Chinese Remainder Theorem?

To solve the previous problem, write out the numbers x ≡ 3 (mod 5) until you find a number congruent to 2 (mod 4) , then increase that number by multiples of 5 x 4 until you find number congruent to 1 (mod 3) . x ≡ 3 (mod 5): x = 3,8,13, 18 ≡ 2 (mod 4) [increase by 5] x ≡ 18 (mod 20): x = 18,38, 58 ≡ 1 (mod 3) [increase by 5 x 4] x ≡ 58 (mod 60).

How is the remainder theorem applied to moduli?

The Chinese remainder theorem can be applied to systems with moduli that are not co-prime, but a solution to such a system does not always exist. { x ≡ 5 ( m o d 6) x ≡ 3 ( m o d 8). ≡ 5 (mod 6) ≡ 3 (mod 8). Note that the greatest common divisor of the moduli is 2. The first congruence implies x ≡ 1 ( m o d 2). x \\equiv 1 \\pmod {2}. x ≡ 1 (mod 2).

How to calculate the modular exponentiation for RSA?

RSA calculations When we come to decrypt ciphertext c(or generate a signature) using RSA with private key (n, d), we need to calculate the modular exponentiation m = cdmod n. The private exponent dis not as convenient as the public exponent, for which we can choose a value with as few ‘1’ bits as possible.

How to find the algorithm for RSA with CRT?

Just give M P and M Q (and P and Q) to the CRT algorithm and you get M. The algorithm you always find when you’re searching for “RSA with CRT” is more complicated than that, you have additional values to compute like q inv and h etc… That’s what you find in Wikipedia and in poncho’s answer.