RSA system with p=17, q=11 and e=3. Find m corresponding to c=156

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SUMMARY

The discussion focuses on solving an RSA decryption problem with parameters p=17, q=11, e=3, and c=156. The official solution calculates the private key d as 107 using the formula d=e^(-1) mod 160. The decryption process involves computing mp and mq using the formula mp = c^d mod p and mq = c^d mod q, both yielding a result of 7. The user seeks clarification on the decryption steps and the relevance of Fermat's Little Theorem in this context.

PREREQUISITES
  • Understanding of RSA encryption and decryption principles
  • Familiarity with modular arithmetic
  • Knowledge of Fermat's Little Theorem
  • Experience with exponentiation by repeated squaring
NEXT STEPS
  • Study the application of Fermat's Little Theorem in RSA decryption
  • Learn about the Chinese Remainder Theorem and its role in RSA
  • Explore the method of exponentiation by repeated squaring
  • Practice RSA decryption problems with different parameters
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Cryptography students, mathematicians, and software developers interested in understanding RSA encryption and decryption processes.

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Homework Statement
Consider an RSA system with p=17, q=11 and e=3.

Find m corresponding to c=156.
Relevant Equations
C = P^e mod n

P = C^d mod n

n = pq

m = (p-1)(q-1)

1 < e < m

d = e^(-1) mod m

1 < d < m
OFFICIAL SOLUTION:
d=e^(-1) mod 160=107

mp= c^(d) mod p=7

mq:=c^(d) mod q=7

MY THOUGHTS:
I understand how d = 107, but I got that by using m = (17-1)(11-1) = 160.

What I don't understand is the next two lines (from the official solution). I am aware of the P = C^d mod n (decryption) formula. Are the aforementioned two lines somehow using "Fermat's little theorem"?

Put differently, I guess the part of the problem statement that I'm struggling with is the "corresponding to c=156" part.

Could someone please help me fully understand how to solve this problem?

Any input would be GREATLY appreciated!

P.S.
I have another problem that's the same problem, except it requires that the solution be found using the Chinese Remainder Theorem, so I'm supposed to do this particular problem without the Chinese Remainder Theorem.
 
Physics news on Phys.org
You can calculate c^d mod p or c^d mod q using exponentiation by repeated squaring. Use the iterative example in the wiki article, except there no need to deal with negative exponent (since d is positive), and the math operations would be performed mod p or mod q.

https://en.wikipedia.org/wiki/Exponentiation_by_squaring
 

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