The paper High-speed high-security signatures by Bernstein et al. introduces the Edwards curve Ed25519.

Concerning the base point $B$, it says that

$B$ is the unique point $(x,4/5)∈E$ for which $x$ is positive

and $B$ corresponds to the basepoint used on the birationally equivalent curve Curve25519.

So I haven't found on any site information about how do they get that $4/5$ value of the $y$ coordinate in order to get a small basepoint generator.

Is there any reason or origin of that $4/5$? How they determined that $4/5$ was part of the group $2^{255} -19$?

  • $\begingroup$ related Base point in Ed25519? $\endgroup$
    – kelalaka
    Mar 14, 2019 at 23:06
  • $\begingroup$ It's related but it doesn't answer my question. Thanks anyway! It's interesting what it's mentioned there $\endgroup$
    – CPereez19
    Mar 14, 2019 at 23:12

1 Answer 1


Over the field $\mathbb Z/(2^{255} - 19)\mathbb Z$, Curve25519 is the Montgomery curve $$v^2 = u^3 + 486662 u^2 + u,$$ and edwards25519 is the twisted Edwards curve $$-x^2 + y^2 = 1 - \frac{121665}{121666} x^2 y^2.$$ The curves correspond by the birational map

\begin{gather} x = \sqrt{-486664} \frac{u}{v}, \quad y = \frac{u - 1}{u + 1}; \\ u = \frac{y + 1}{y - 1}, \quad v = \sqrt{-486664} \frac{u}{x}. \end{gather}

The standard base point $P$ for the X25519 Diffie–Hellman function has $u(P) = 9$; the $v$ coordinate is not significant because X25519 works exclusively with $u$ coordinates, so there are two possible points, $P = (9, \pm \sqrt{9^3 + 486662\cdot 9^2 + 9})$. The choice of $u(P)$ is arbitrary as long as it corresponds to a point of large prime order—it can't affect security—but was made to be the smallest such integer. The standard base point $Q$ for Ed25519 signature was chosen to correspond to one of the possible choices for $P$, so $$y(Q) = \frac{u(P) - 1}{u(P) + 1} = \frac{9 - 1}{9 + 1} = 4/5;$$ then the $x$ coordinate of $Q$ was chosen arbitrarily from the two options.


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