drewfstr314
- 20
- 0
If one is solving a modular equation:
4k \equiv 1 \: (\text{mod } n)
with n even, known, for k, then one needs to find the inverse of 4 modulo n:
4x - 1 = nc
4x - nc = 1
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!
4k \equiv 1 \: (\text{mod } n)
with n even, known, for k, then one needs to find the inverse of 4 modulo n:
4x - 1 = nc
4x - nc = 1
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!