- #1
drewfstr314
- 20
- 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!
[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!