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I'm working on Shamir secret sharing and i came across a part where i have

$(-4)^{-1}(\mod 17)$

I have never done negative number using inverse modulus so im kinda stuck

i tried changing it to

$(13)^{-1}(\mod 17)$

but it doesn't seem to work.

Gerry Myerson
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txr112
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    You've started off fine, and 13 does have an inverse modulo 17 since 13 and 17 are coprime. How have you attempted to find it? Having said that, it might be easier to recognize that $-16\equiv 1\pmod{17}$ – Casteels Feb 28 '14 at 10:46
  • ok i think i made a careless mistake along the way resulting in me not getting the value correctly. now it works out fine. thanks for assuring me that im on the right track! – txr112 Feb 28 '14 at 11:12

1 Answers1

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It is actually better to keep $-4$, since it's smaller than $13$ and computations will be easier. Since $(-1)^{-1}=-1$, it suffices you compute $4^{-1}$. But $4\times (-4)=-16\equiv 1$. So the inverse of $-4$ will be...?

Pedro
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