Question

Bob and Alice want to choose a key they can use for cryptography, but all they have to communicate is a bugged phone line. Bob proposes that they each choose a secret number, $a$ for Alice and $b$ for Bob. They also choose, over the phone, a prime number $p$ with more digits than any key they want to use, and one more number $q$. Bob will send Alice $b q$ $(\bmod p)$, and Alice will send Bob aq $(\bmod p)$. Their key (which they will keep secret) will then be $a b q(\bmod p)$. (Here we don't worry about the details of how they use their key, only with how they choose it.) As Bob explains, their wire tapper will know $p, q, a q$ $(\bmod p)$, and $b q(\bmod p)$, but will not know $a$ or $b$, so their key should be safe. Is this scheme safe, that is can the wiretapper compute $a b q \bmod p$ ? If so, how does she do it. Alice says "You know, the scheme sounds good, but wouldn't it be more complicated for the wire tapper if I send you $q^a \quad(\bmod p)$, you send $\operatorname{me} q^b \quad(\bmod p)$ and we use $q^{a b} \quad(\bmod p)$ as our key?" In this case can you think of a way for the wire tapper to compute $q^{a b}$ $(\bmod p)$. If so, how can you do it? If not, what is the stumbling block? (It is fine for the stumbling block to be that you don't know how to compute something, you don't need to prove that you can't compute it.)

   Bob and Alice want to choose a key they can use for cryptography, but all they have to communicate is a bugged phone line. Bob proposes that they each choose a secret number, $a$ for Alice and $b$ for Bob. They also choose, over the phone, a prime number $p$ with more digits than any key they want to use, and one more number $q$. Bob will send Alice $b q$ $(\bmod p)$, and Alice will send Bob aq $(\bmod p)$. Their key (which they will keep secret) will then be $a b q(\bmod p)$. (Here we don't worry about the details of how they use their key, only with how they choose it.) As Bob explains, their wire tapper will know $p, q, a q$ $(\bmod p)$, and $b q(\bmod p)$, but will not know $a$ or $b$, so their key should be safe.
Is this scheme safe, that is can the wiretapper compute $a b q \bmod p$ ? If so, how does she do it.
Alice says "You know, the scheme sounds good, but wouldn't it be more complicated for the wire tapper if I send you $q^a \quad(\bmod p)$, you send $\operatorname{me} q^b \quad(\bmod p)$ and we use $q^{a b} \quad(\bmod p)$ as our key?" In this case can you think of a way for the wire tapper to compute $q^{a b}$ $(\bmod p)$. If so, how can you do it? If not, what is the stumbling block? (It is fine for the stumbling block to be that you don't know how to compute something, you don't need to prove that you can't compute it.)
Show more…
Discrete Math in Computer Science
Discrete Math in Computer Science
Bogart, Stein.
Chapter 2, Problem 3 ↓

Instant Answer

verified

Step 1

They also publicly agree on a prime number $p$ and another number $q$. Alice sends $aq \pmod{p}$ to Bob, and Bob sends $bq \pmod{p}$ to Alice. Their shared secret key would then be $abq \pmod{p}$.  Show more…

Show all steps

lock
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
Bob and Alice want to choose a key they can use for cryptography, but all they have to communicate is a bugged phone line. Bob proposes that they each choose a secret number, $a$ for Alice and $b$ for Bob. They also choose, over the phone, a prime number $p$ with more digits than any key they want to use, and one more number $q$. Bob will send Alice $b q$ $(\bmod p)$, and Alice will send Bob aq $(\bmod p)$. Their key (which they will keep secret) will then be $a b q(\bmod p)$. (Here we don't worry about the details of how they use their key, only with how they choose it.) As Bob explains, their wire tapper will know $p, q, a q$ $(\bmod p)$, and $b q(\bmod p)$, but will not know $a$ or $b$, so their key should be safe. Is this scheme safe, that is can the wiretapper compute $a b q \bmod p$ ? If so, how does she do it. Alice says "You know, the scheme sounds good, but wouldn't it be more complicated for the wire tapper if I send you $q^a \quad(\bmod p)$, you send $\operatorname{me} q^b \quad(\bmod p)$ and we use $q^{a b} \quad(\bmod p)$ as our key?" In this case can you think of a way for the wire tapper to compute $q^{a b}$ $(\bmod p)$. If so, how can you do it? If not, what is the stumbling block? (It is fine for the stumbling block to be that you don't know how to compute something, you don't need to prove that you can't compute it.)
Close icon
Play audio
Feedback
Powered by NumerAI
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Join the community

18,000,000+

Students on Numerade


Trusted by students at 8,000+ universities

Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever