Is the STARK Curve a SafeCurve?

The STARK Curve seems a reasonable choice for ECDSA.

The STARK Curve defined over $\mathbb{F}_p$ with $p = 2^{251} + 17*2^{192} +1$ with the short Weierstrass equation

$$y^2 = x^3 + A x + B$$

with

• $A = 1$, and
• $B = 3141592653589793238462643383279502884197169399375105820974944592307816406665$

• The order of the curve group ( numbner of points) is $n = \#E(\mathbf{F}_p )$ is $n= 3618502788666131213697322783095070105526743751716087489154079457884512865583$

  And this is a prime number indication that

  • Every element except the identity ( $\mathcal{O})$ can be a generator. The nothing-in-my-sleeve number of this curve (thanks to Aria for pointing), comes from the $\pi$.

    So Starks has Somewhat rigidness at least for now.

    In the end, the nothing-in-my-sleeve number is rather physiological.

  • Co-factor is $h=1$ this means that there is no Montgomery representation of the curve, as a result, there is no fast Montgomery ladder (requires an element of order 2, i.e. 2|co-factor), Joyce ladder is still possible with slower performance. In ECDSA this is helpful in the calculation of $[k]G$ since only $x$ coordinate is used.

  • There is no small group attack to consider, though this is not a problem for legitimate users of ECDSA. If the users are not legitimate then they can use this to double-spend coins as did in Curv25519 however this is not the case for the STARK curve.

  • The curve group is isomorphic to $\mathbb{Z_n}$

• The $n$ has 252-bit binary representation and this implies it has around $126$-bit security against the best classical discrete logarithm problems.

• The size of the curve gives no collision of $k$ if a good random number generator is used. If one still fears this one can use deterministic ECDSA given in rfc-6979.

• Twist security ( not related to ECDSA); the quadratic twist of this curve is $$y^2 = x^3 + 5^2*x +B*5^3$$ ^*

  • The cardinality of the twist = "618502788666131213697322783095070105623107215331596699973092056135872020481"
  • The factors of the twist group = "499669 * 26023817775804638430931 * 278275836047110893120702478691334736277272165979" and this gives around 158 bit security. Moderate level.

• And, we have $2*p+2 = Ord(E) + Ord(\text{E_quaratic_twist})$

• $n \neq p$ therefore it is not an anomalous curve where the discrete log can be solved quickly.

a = 1
b = 3141592653589793238462643383279502884197169399375105820974944592307816406665
p = 2^251 + 17*2^192 +1

E = EllipticCurve(GF(p), [0,0,0,a,b])

print(E)
Et = E.quadratic_twist()
print(Et)

print("E abelian =", E.abelian_group())
print("E twist a = ", Et.abelian_group())

card = E.cardinality()
cardEt = Et.cardinality()

print("cardinality E       =",card)
print("cardinality E twist =",card)


print("factors E   ",factor(card))
print("factors Et ",factor(cardEt))

#Generator part not for the quadratic twist.
#G = E(874739451078007766457464989774322083649278607533249481151382481072868806602,152666792071518830868575557812948353041420400780739481342941381225525861407)
#n = G.order()
#print("Generator order =", n)

print(log(card,2).n()+1)

assert(2*p+2 == card + cardEt)

_{^*The quadratic twist formed with QNR 5, unfortunately, it did not work as intended. Thanks to Poncho to point out this. I keep the equation so that one can see the problem. Instead, I've used quadratic_twist function of SageMath that quite slow.}