I am trying to convert $(g^a)^{i^2}\mod p$ to an equivalent Elliptic Curve expression. Let's assume a base point $G$ over an appropriate ECC. Will $(g^a)^{i^2}\mod p$ be equivalent to $G\cdot a\cdot i^2$ (where the dot operation is multiplication over the elliptic curve)?

  • $\begingroup$ Yes it will be. $\endgroup$
    – SEJPM
    Aug 6, 2018 at 11:17
  • $\begingroup$ Ok, thanks. Now, is there a source which explains why this is true? $\endgroup$
    – Shak
    Aug 6, 2018 at 11:24

1 Answer 1


So this is more a question of notation than anythig else. So let's examine the notation.

$$\left(g^{a}\right)^{i^2}\bmod p$$

which is equivalent to

$$g^{a\cdot i^2}\bmod p$$

means the following: Consider the group $\mathbb Z _p^*$, that is the set of non-negative integers smaller than $p$ along with the neutral element $1$ and the group operation of multiplication modulo $p$ and write this group multiplicatively with $g\in\mathbb Z_p^*$ and $a,i$ being normal integers.

Now if, instead we wrote the above as an additive group (which is merely a change of notation!), we would replace $\cdot$ with $+$ and exponentiation $g^x$ with multiplication by a scalar $x\cdot G$. So the above expression immediately becomes $(a\cdot i^2)\cdot G$ (up to commutativity and associativity). And the commutativity is a given on elliptic curves (because otherwise DH key exchanges wouldn't work).


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