Skip to main content
MathOverflow

Return to Question

Post Closed as "Not suitable for this site" by Emil Jeřábek, Ben Barber, user44191, Felipe Voloch, Max Alekseyev
Math Jaxed+ minor typo removal
Source Link
edit approved Jan 28, 2020 at 14:10
Daniele Tampieri
edit approved Jan 28, 2020 at 14:10
Daniele Tampieri
  • 6.4k
  • 7
  • 30
  • 45

Suppose I have an arbitrary 256 bit number m$m$ another secret number k$k$ of the same bit length, and then I multiply them both modulo a 256 bit prime number p$p$ to get c$c$ as follows: c = (m*k) mod p$$ c = (m\cdot k) \mod p $$ Is there any way to get m$m$ back without knowing k $k$? Is this problem as hard as the descretediscrete log problem? How can this task be made more computationally difficult?

Suppose I have an arbitrary 256 bit number m another secret number k of the same bit length, and then I multiply them both modulo a 256 bit prime number p to get c as follows: c = (m*k) mod p Is there any way to get m back without knowing k ? Is this problem as hard as the descrete log problem? How can this task be made more computationally difficult?

Suppose I have an arbitrary 256 bit number $m$ another secret number $k$ of the same bit length, and then I multiply them both modulo a 256 bit prime number $p$ to get $c$ as follows: $$ c = (m\cdot k) \mod p $$ Is there any way to get $m$ back without knowing $k$? Is this problem as hard as the discrete log problem? How can this task be made more computationally difficult?

Source Link
Aravind A
Aravind A
  • 117
  • 5

Is there any way to solve this equation without knowing the inverse modulo?

Suppose I have an arbitrary 256 bit number m another secret number k of the same bit length, and then I multiply them both modulo a 256 bit prime number p to get c as follows: c = (m*k) mod p Is there any way to get m back without knowing k ? Is this problem as hard as the descrete log problem? How can this task be made more computationally difficult?