# Modular Arithmetic and Diophantine Equations

1. Oct 25, 2012

### drewfstr314

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!

2. Oct 25, 2012

### tiny-tim

hi drewfstr314!

that's right

4k is even, and 1 (mod n) is odd (since n is even), so there's no solution …

what is worrying you about that?