Take the 2-minute tour ×
Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

You can see a little background about this on this bitcointalk post by the late Hal Finney.

$\beta$ and $\lambda$ are the values on the secp256k1 curve such that: $$\begin{align} \lambda^3 &= 1 \mod N \\ \beta^3 &= 1 \mod P \\ \end{align}$$

As seen here, in hex, $N$ and $P$ are: $$\begin{align} N &= \mathtt{FFFFFFFF\ FFFFFFFF\ FFFFFFFF\ FFFFFFFE\ BAAEDCE6\ AF48A03B\ BFD25E8C\ D0364141} \\ P &= \mathtt{FFFFFFFF\ FFFFFFFF\ FFFFFFFF\ FFFFFFFF\ FFFFFFFF\ FFFFFFFF\ FFFFFFFE\ FFFFFC2F} \\ \end{align}$$

The actual values of lambda and beta are easily verifiable and are: $$\begin{align} \lambda &= \mathtt{5363ad4c\ c05c30e0\ a5261c02\ 8812645a\ 122e22ea\ 20816678\ df02967c\ 1b23bd72} \\ \beta &= \mathtt{7ae96a2b\ 657c0710\ 6e64479e\ ac3434e9\ 9cf04975\ 12f58995\ c1396c28\ 719501ee} \\ \end{align}$$

The question for me is, how do you derive this? Can someone show me step-by-step how you can figure out these values?

Reposted from Bitcoin Stack Exchange

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Given that $N$ and $P$ are prime, one obvious way to do this is to select a random value $g$ from $[1, N-1]$, and compute $g^{(N-1)/3} \bmod N$; assuming that $N \equiv 1 \pmod{3}$, this resulting value will either be 1, the displayed value of $\lambda$, or $N-\lambda-1$ (with equal probabilities of each). If $N \not\equiv 1 \pmod{3}$, then the only modular cube root of 1 will be 1.

And, to compute $\beta$, you do the same with $P$.

share|improve this answer
Thank you so much! I've confirmed this works for both $\lambda$ and $\beta$. What number theory theorem is used to come up with this? –  jimmysong Feb 3 at 4:53
Actually, nevermind, it's clear you're using Fermat's little theorem. –  jimmysong Feb 3 at 5:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.