Modular Arithmetic and Diophantine Equations

  • Context: Undergrad 
  • Thread starter Thread starter drewfstr314
  • Start date Start date
  • Tags Tags
    Arithmetic
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 3K views
drewfstr314
Messages
20
Reaction score
0
If one is solving a modular equation:

[itex]4k \equiv 1 \: (\text{mod } n)[/itex]

with n even, known, for k, then one needs to find the inverse of 4 modulo n:

[itex]4x - 1 = nc[/itex]
[itex]4x - nc = 1[/itex]

But this only has solutions iif (4,n) = 2 (n is even, but not a multiple of 4), which doesn't divide 1, so there is no inverse of 4 modulo n. Does this mean that there isn't a k that satisfies the original equation?

Thanks!
 
Physics news on Phys.org