If one is solving a modular equation:(adsbygoogle = window.adsbygoogle || []).push({});

[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!

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# Modular Arithmetic and Diophantine Equations

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