Bob decides to use $n = 697 \rightarrow 17 × 41$ and $e = 33$ as his public key for an RSA cryptosystem.

  1. Show that the decryption exponent is $97$.

  2. Find the encrypted form of the message $17$.


Since you "have done a bit of work on this", it would be nice to actually show that "bit of work".

Anyway, here are three hints:

  1. The exercise phrasing is actually wrong. $97$ is not "the" decryption exponent. It's only "a" decryption exponent, among a set of decryption exponents that match the public exponent $e = 33$. That set is infinite. $97$ is not even the smallest value in that set of solution; it is just the "traditional" solution. Another solution is $17$. In fact, all integers $17+80k$ for any $k\ge 0$ are "private exponents" that match $e = 33$ for modulus $n = 697$.

  2. Everything in RSA can be done with the Chinese Remainder Theorem. In a nutshell, that theorem says that since $41$ and $17$ are prime to each other, all computations modulo $41\times 17 = 697$ are equivalent to computing things modulo $41$ and modulo $17$ in parallel. This means that you can split your problem of verifying that $97$ is a proper private exponent into two sub-problems: verifying that it works modulo $41$, and verifying that it works modulo $17$. The CRT really tells you that if it works modulo both, then it will work modulo their product $697$.

  3. For the second question, use the CRT again.

With these hints, you can solve all of it with a pen-and-paper, or even doing the computations in your head (I know it's feasible, I just did).

  • $\begingroup$ Here $\phi(n) = 16*40 = 640$. and 17*33 = 561. How is it a solution? Why is $17+80k$ a solution if $\phi(n) \neq 80$? $\endgroup$ – satya Dec 12 '18 at 11:17
  • 1
    $\begingroup$ @satya It is not necessary that $d$ is an inverse of $e$ modulo $\phi(n)$. For RSA to work, you just need to have $d$ an inverse of $e$ modulo both $p-1$ and $q-1$ (the CRT tells you this is sufficient; try it!). For that, you make $d$ and inverse of $e$ modulo the least common multiple of $p-1$ and $q-1$, which is 80 in that case. The Euler function $\phi(n) = (p-1)(q-1)$ is a multiple of that common multiple. $\endgroup$ – Thomas Pornin Dec 12 '18 at 15:16

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