Normally I use the naive method: $$a^{-1} = a \cdot b \bmod p \equiv 1,$$ where b is the inverse of a.

Else I love to use Fermat's little theorem: $$a^{p − 1} \equiv 1 \bmod p.$$ By multiplying both sides with $a^{-1}$ you get, that the inverse is $a^{p-2}$.

But let us say I have a field of the size 8 $(F_8)$ and I shall find the inverse of $t+2$, how do I do?

  • $\begingroup$ What is $t$ in $\mathbb F_8$? $\endgroup$ – lhf May 26 at 21:47
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    $\begingroup$ By the way, you should notice that, in $\Bbb F_8$, $t+2=t$. $\endgroup$ – Gae. S. May 26 at 21:51
  • $\begingroup$ Euclidean algorithm in the Bezout lemma form $\endgroup$ – crystal_math May 26 at 21:54

In $\mathbb F_8$, we have $2=0$.

Assuming that $t+2\ne0$, we have $(t+2)^{-1} = t^{-1} = t^6$, because $\mathbb F_8^\times$ has order $7$.

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