Science:Math Exam Resources/Courses/MATH312/December 2008/Question 05 (b)

MATH312 December 2008

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[hide] Question 05 (b)

Alice wants Bob to send her a secret integer P between 0 and 1250 using RSA encryption with key $\displaystyle (e,n)=(1189,1271)$ (so the exponent is 1189 and the modulus is 1271, it is way too small to be really secure, but basic RSA encryption and decryption methods still work of course). (b) To obtain P from C, Alice first computes an inverse $\displaystyle d\in \mathbb {Z}$ of e modulo $\displaystyle \phi (n)$, since she created the key, she knows that $\displaystyle 1271=31\cdot 41$ (in fact, the prime factorization of n can easily be found since it is so small). Compute such a d.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Euclidean algorithm to the rescue! Do this with the number $\displaystyle \phi (n)$ and 1189 and then back substitute.

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[show] Solution

We use the Euclidean algorithm to find an inverse. We do this with the numbers 1189 and $\displaystyle \phi (n)=\phi (1271)=\phi (31)\phi (41)=30\cdot 40=1200$. This gives $\displaystyle {\begin{aligned}1200&=1189(1)+11\\1189&=11(108)+1\end{aligned}}$ and back substituting gives $\displaystyle {\begin{aligned}1&=1189-11(108)\\1&=1189-(1200-1189(1))(108)\\1&=1189(109)-1200(108)\\\end{aligned}}$ Hence the value of d can be chosen to be $\displaystyle d=109$ since $\displaystyle 1189(109)\equiv 1\mod {\phi (1271)}$.

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