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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

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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 $P$.

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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

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