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Python code to get beta and lambda values for p and n of secp256k1 curve

Getting beta of p

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, (p-1)/3, p)

beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

Getting lambda of n

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, (n-1)/3 , n)

lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72


More info

I experimented further with this by getting beta and lambda for both p and n and discovered that all the results generated become useful for finding the identical values for x or y in the equation y ^ 2 = x ^ 3 + 7 mod p

#beta and lambda for p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, (p-1)/3, p)
lambdaOfP = pow(3, (p-1)/3, p)
print "betaOfP \t= 0x%x " % betaOfP 
print "lambdaOfP\t= 0x%x " % lambdaOfP 
print

#beta and lambda for n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, (n-1)/3 , n)
lambdaOfN = pow(3, (n-1)/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN

betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

 

betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

Python code to get beta and lambda values for p and n of secp256k1 curve

Getting beta of p

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, (p-1)/3, p)

beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

Getting lambda of n

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, (n-1)/3 , n)

lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72


More info

I experimented further with this by getting beta and lambda for both p and n and discovered that all the results generated become useful for finding the identical values for x or y in the equation y ^ 2 = x ^ 3 + 7 mod p

#beta and lambda for p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, (p-1)/3, p)
lambdaOfP = pow(3, (p-1)/3, p)
print "betaOfP \t= 0x%x " % betaOfP 
print "lambdaOfP\t= 0x%x " % lambdaOfP 
print

#beta and lambda for n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, (n-1)/3 , n)
lambdaOfN = pow(3, (n-1)/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN

betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

 

betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

Python code to get beta and lambda values for p and n of secp256k1 curve

Getting beta of p

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, (p-1)/3, p)

beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

Getting lambda of n

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, (n-1)/3 , n)

lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72


More info

I experimented further with this by getting beta and lambda for both p and n and discovered that all the results generated become useful for finding the identical values for x or y in the equation y ^ 2 = x ^ 3 + 7 mod p

#beta and lambda for p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, (p-1)/3, p)
lambdaOfP = pow(3, (p-1)/3, p)
print "betaOfP \t= 0x%x " % betaOfP 
print "lambdaOfP\t= 0x%x " % lambdaOfP 
print

#beta and lambda for n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, (n-1)/3 , n)
lambdaOfN = pow(3, (n-1)/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN

betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

my ocd
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Python code to get beta and lambda values for p and n of secp256k1 curve

Getting beta of p

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, (p-1)/3, p)

beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

Getting lambda of n

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, (n-1)/3 , n)

lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72


More info

I experimented further with this by getting beta and lambda for both p and n and discovered that all the results generated become useful for finding the identical values for x or y in the equation y ^ 2 = x ^ 3 + 7 mod p

#powers#beta ofand cubedlambda for p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, (p-1)/3, p)
lambdaOfP = pow(3, (p-1)/3, p)
print "betaOfP \t= 0x%x " % betaOfP 
print "lambdaOfP\t= 0x%x " % lambdaOfP 
print

#powers#beta ofand cubedlambda for n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, (n-1)/3 , n)
lambdaOfN = pow(3, (n-1)/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN

betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

Python code to get beta and lambda values for p and n of secp256k1 curve

Getting beta of p

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, (p-1)/3, p)

beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

Getting lambda of n

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, (n-1)/3 , n)

lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72


More info

I experimented further with this by getting beta and lambda for both p and n and discovered that all the results generated become useful for finding the identical values for x or y in the equation y ^ 2 = x ^ 3 + 7 mod p

#powers of cubed p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, (p-1)/3, p)
lambdaOfP = pow(3, (p-1)/3, p)
print "betaOfP \t= 0x%x " % betaOfP 
print "lambdaOfP\t= 0x%x " % lambdaOfP 
print

#powers of cubed n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, (n-1)/3 , n)
lambdaOfN = pow(3, (n-1)/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN

betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

Python code to get beta and lambda values for p and n of secp256k1 curve

Getting beta of p

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, (p-1)/3, p)

beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

Getting lambda of n

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, (n-1)/3 , n)

lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72


More info

I experimented further with this by getting beta and lambda for both p and n and discovered that all the results generated become useful for finding the identical values for x or y in the equation y ^ 2 = x ^ 3 + 7 mod p

#beta and lambda for p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, (p-1)/3, p)
lambdaOfP = pow(3, (p-1)/3, p)
print "betaOfP \t= 0x%x " % betaOfP 
print "lambdaOfP\t= 0x%x " % lambdaOfP 
print

#beta and lambda for n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, (n-1)/3 , n)
lambdaOfN = pow(3, (n-1)/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN

betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

made my equations more correct
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Python code to get beta and lambda values for p and n of secp256k1 curve

Getting beta of p

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, (p-1)/3, p)

beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

Getting lambda of n

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, (n-1)/3 , n)

lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72


More info

While I don't understand the true meaning of 'beta' and 'lambda', it seems it is related to the number which is the base number of the power operation. So I experimented further and discovered that if I producewith this by getting beta and lambda for both p and n, and discovered that all the results generated become useful for finding the identical values offor x or y in the equation y ^ 2 = x ^ 3 + 7 mod p

#powers of cubed p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, (p-1)/3, p)
lambdaOfP = pow(3, (p-1)/3, p)
print "betaOfP \t= 0x%x " % betaOfP 
print "lambdaOfP\t= 0x%x " % lambdaOfP 
print

#powers of cubed n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, (n-1)/3 , n)
lambdaOfN = pow(3, (n-1)/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN

betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

Python code to get beta and lambda values for p and n of secp256k1 curve

Getting beta of p

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, p/3, p)

beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

Getting lambda of n

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, n/3 , n)

lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72


More info

While I don't understand the true meaning of 'beta' and 'lambda', it seems it is related to the number which is the base number of the power operation. So I experimented further and discovered that if I produce beta and lambda for both p and n, all the results generated become useful for finding the values of x or y in the equation y ^ 2 = x ^ 3 + 7 mod p

#powers of cubed p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, p/3, p)
lambdaOfP = pow(3, p/3, p)
print "betaOfP \t= 0x%x " % betaOfP 
print "lambdaOfP\t= 0x%x " % lambdaOfP 
print

#powers of cubed n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, n/3 , n)
lambdaOfN = pow(3, n/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN

betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

Python code to get beta and lambda values for p and n of secp256k1 curve

Getting beta of p

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, (p-1)/3, p)

beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

Getting lambda of n

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, (n-1)/3 , n)

lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72


More info

I experimented further with this by getting beta and lambda for both p and n and discovered that all the results generated become useful for finding the identical values for x or y in the equation y ^ 2 = x ^ 3 + 7 mod p

#powers of cubed p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, (p-1)/3, p)
lambdaOfP = pow(3, (p-1)/3, p)
print "betaOfP \t= 0x%x " % betaOfP 
print "lambdaOfP\t= 0x%x " % lambdaOfP 
print

#powers of cubed n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, (n-1)/3 , n)
lambdaOfN = pow(3, (n-1)/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN

betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72

simplified equations
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my ocd
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added a bit more background info
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