I'm learning the multiplication operation on EC. From most material I can found, the multiplication $nP$ is just:

$$nP=P+P+\cdots +P+P$$ For negative factor, i.e. $(-n)P$, by above definition and the definition of $-P$, we can write $(-n)P$ as:

$$(-n)P = (-P) + (-P) + \cdots (-P)$$

My question is: is it possible to expression $(-n)P$ as $mP$ where $m>0$ under reasonable assumption? For example, we can assume the knowledge of the group order.

I'm asking so because $nP$ on EC is analogous to exponentiation in group $\mathbb{G}$ with order $p$. And in the latter case we have $g^{-n} \equiv g ^{k\phi(q)-n}\pmod q, \forall k\in \mathbb{Z}$ by Fermat's little. So I'm wondering if there is a similar result for EC.


1 Answer 1


If you know the order of the curve (we'll call it $q$), then it's easy.

In that case, we have the identity $xP = (x + q)P$ for any $x$. Using that identity, we can express $(-n)P$ as $(q - n)P$; by the above identity, these are the same. If $q > n$, this gives you your answer.

The more general answer (which works for any $n$) is $(-n \bmod q)P$, using the convention that $a \bmod b$ is that integer which is between 0 and $b-1$, and which is congruent to $a$ modulo $b$.

This is exactly analogous to the $\mathbb{G}^*$ case; in that case, the size of the group is $\phi(q)$; in this Elliptic curve case, the size of the group is $q$.


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