Arranged
5 minutes to read
We are given the SageMath source code to encrypt the flag:
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from Crypto.Util.number import long_to_bytes
from hashlib import sha256
from secret import FLAG, p, b, priv_a, priv_b
F = GF(p)
E = EllipticCurve(F, [726, b])
G = E(926644437000604217447316655857202297402572559368538978912888106419470011487878351667380679323664062362524967242819810112524880301882054682462685841995367, 4856802955780604241403155772782614224057462426619061437325274365157616489963087648882578621484232159439344263863246191729458550632500259702851115715803253)
A = G * priv_a
B = G * priv_b
print(A)
print(B)
C = priv_a * B
assert C == priv_b * A
# now use it as shared secret
secret = C[0]
hash = sha256()
hash.update(long_to_bytes(secret))
key = hash.digest()[16:32]
iv = b'2__\xd9]3k\x94\x893\x1a\x7f\x93\xd5\x14\x05'
cipher = AES.new(key, AES.MODE_CBC, iv)
encrypted = cipher.encrypt(pad(FLAG, 16))
print(encrypted)
Moreover, we have the output of the script:
(6174416269259286934151093673164493189253884617479643341333149124572806980379124586263533252636111274525178176274923169261099721987218035121599399265706997 : 4355483362759117096840903200308104416093873813961514429031587444212486652000030486511593222701883061189598478367449346932543603997532721961479304197729837 : 1)
(5988601072335190267607626269694202717170383512532989213900345448657641222212196937725485374210603546742409649154478860998573202499999124529884522786776964 : 1979660325653723467116529048392433203266542286102635118367760228627475646827461168434739881080159207311102652868018376000583885500682322201011066971591840 : 1)
b'\xe3Nf1\xd4\xf7g\xad\xa6\xa4q\x8e\x85\x99\xa2/>\xb0u\x16\x1f\xc5\x1e\x8a\xf3\xd0t\xf5\xc4F\x9a\xce'
Source code analysis
The server uses ECC for a Diffie-Hellman key exchange (ECDH). It is using a custom elliptic curve, wich unknown parameters $p$ and $b$ (notice that $a = 726$), so
$$ E(\mathbb{F}_p): y^2 = x^3 + 726 \cdot x + b \mod{p} $$
The server takes a known generator point $G$ and uses private numbers $\mathrm{priv\_a}$ and $\mathrm{priv\_b}$ to compute $A = \mathrm{priv\_a} \cdot G$ and $B = \mathrm{priv\_b} \cdot G$. We are given these points $A$ and $B$ (and we know $G$ from the source code).
After that, the shared secret is just the $\mathrm{x}$ coordinate of $C = \mathrm{priv\_a} \cdot B$, which is also equal to $\mathrm{priv\_b} \cdot A$. That’s how ECDH works, in general terms.
Solution
The first thing we need to operate with the points in the curve is to actually find the curve parameters $b$ and $p$.
Finding curve parameters
Notice that we already have three points that lie on the curve, so we have three equations that hold:
$$ \begin{cases} (G_\mathrm{y})^2 = (G_\mathrm{x})^3 + 726 \cdot G_\mathrm{x} + b & \mod{p} \\ (A_\mathrm{y})^2 = (A_\mathrm{x})^3 + 726 \cdot A_\mathrm{x} + b & \mod{p} \\ (B_\mathrm{y})^2 = (B_\mathrm{x})^3 + 726 \cdot B_\mathrm{x} + b & \mod{p} \\ \end{cases} $$
If we subtract the three equations to eliminate $b$, we get these two:
$$ \begin{cases} (G_\mathrm{y})^2 - (A_\mathrm{y})^2 = (G_\mathrm{x})^3 - (A_\mathrm{x})^3 + 726 \cdot (G_\mathrm{x} - A_\mathrm{x}) & \mod{p} \\ (G_\mathrm{y})^2 - (B_\mathrm{y})^2 = (G_\mathrm{x})^3 - (B_\mathrm{x})^3 + 726 \cdot (G_\mathrm{x} - B_\mathrm{x}) & \mod{p} \\ \end{cases} $$
If we arrange the equations we have:
$$ \begin{cases} (G_\mathrm{y})^2 - (A_\mathrm{y})^2 - (G_\mathrm{x})^3 + (A_\mathrm{x})^3 - 726 \cdot (G_\mathrm{x} - A_\mathrm{x}) = 0 & \mod{p} \\ (G_\mathrm{y})^2 - (B_\mathrm{y})^2 - (G_\mathrm{x})^3 + (B_\mathrm{x})^3 - 726 \cdot (G_\mathrm{x} - B_\mathrm{x}) = 0 & \mod{p} \\ \end{cases} $$
What we see here is that we have two results that are congruent to $0 \mod{p}$, which means that both results are divisible by $p$. As a result, both results share a common factor $p$, so we can find $p$ using the greatest common devisor (GCD). Well, more precisely, the GCD will output a multiple of $p$, but this time it is just $p$:
$ sage -q
sage: Gx, Gy = (9266444370006042174473166558572022974025725593685389789128881064194700114878783516673806793236640623625249672428198101125248803018820546824626
....: 85841995367, 4856802955780604241403155772782614224057462426619061437325274365157616489963087648882578621484232159439344263863246191729458550632500259702
....: 851115715803253)
sage: Ax, Ay = (6174416269259286934151093673164493189253884617479643341333149124572806980379124586263533252636111274525178176274923169261099721987218035121599
....: 399265706997, 435548336275911709684090320030810441609387381396151442903158744421248665200003048651159322270188306118959847836744934693254360399753272196
....: 1479304197729837)
sage: Bx, By = (5988601072335190267607626269694202717170383512532989213900345448657641222212196937725485374210603546742409649154478860998573202499999124529884
....: 522786776964, 197966032565372346711652904839243320326654228610263511836776022862747564682746116843473988108015920731110265286801837600058388550068232220
....: 1011066971591840)
sage:
sage: a = 726
sage:
sage: p = gcd(
....: Gy ** 2 - Ay ** 2 - Gx ** 3 - a * Gx + Ax ** 3 + a * Ax,
....: Gy ** 2 - By ** 2 - Gx ** 3 - a * Gx + Bx ** 3 + a * Bx,
....: )
sage: is_prime(p)
False
As can be seen, the value of p is not prime, but a small multiple of the prime number we are looking for:
sage: p
40869841224700244502552707772694043870192866604192451956598992316181984174649795098929076591403964203529570112175651477399141927360387112911853903291825198
sage: factor(p)
2 * 3 * 6811640204116707417092117962115673978365477767365408659433165386030330695774965849821512765233994033921595018695941912899856987893397852151975650548637533
sage: p = 6811640204116707417092117962115673978365477767365408659433165386030330695774965849821512765233994033921595018695941912899856987893397852151975650548
....: 637533
Now, it is trivial to find $b$ with a single point:
$$ b = (G_\mathrm{y})^2 - (G_\mathrm{x})^3 - 726 \cdot G_\mathrm{x} \mod{p} $$
sage: b = (Gy ** 2 - Gx ** 3 - a * Gx) % p
Now we can define the elliptic curve and use the above coordinates as points on the curve:
sage: E = EllipticCurve(GF(p), [a, b])
sage: G, A, B = E(Gx, Gy), E(Ax, Ay), E(Bx, By)
ECDLP
Now, we must solve the ECDLP in order to find one of $\mathrm{priv\_a}$ or $\mathrm{priv\_b}$. The first thing we need is find the order of $G$, which will determine if we can solve the ECDLP easily or not:
sage: G.order()
11
Oh! That’s a very small order! This is not good for ECDH, because it means that $11 \cdot G = O$, the “point at infinity” (the identity element on the elliptic curve group).
As a result, no matter how large are $\mathrm{priv\_a}$ or $\mathrm{priv\_b}$, because we can express any of them as $\mathrm{priv\_a} = k \cdot 11 + r$, for some integers $k$ and $r$ with $0 \leqslant r < 11$.
As a result,
$$ \begin{align} A & = \mathrm{priv\_a} \cdot G \\ & = (k \cdot 11 + r) \cdot G \\ & = (k \cdot 11) \cdot G + r \cdot G \\ & = k \cdot (11 \cdot G) + r \cdot G \\ & = k \cdot O + r \cdot G \\ & = O + r \cdot G \\ & = r \cdot G \end{align} $$
The same happens with $B$, so we can simply try different values of $r$ (from $0$ to $10$) and see if $A$ equals $r \cdot G$, and the same with $B$:
sage: for priv_a in range(11):
....: if priv_a * G == A:
....: break
....:
sage: for priv_b in range(11):
....: if priv_b * G == B:
....: break
....:
sage: C = priv_a * B
sage: C == priv_b * A
True
We have got the shared secret, so we can decrypt the flag with AES and the key derived from the shared secret.
Flag
Here’s the flag:
sage: from Crypto.Cipher import AES
sage: from Crypto.Util.Padding import unpad
sage: from Crypto.Util.number import long_to_bytes
sage: from hashlib import sha256
sage:
sage: secret = int(C[0])
sage:
sage: h = sha256()
sage: h.update(long_to_bytes(secret))
sage:
sage: key = h.digest()[16:32]
sage: iv = b'2__\xd9]3k\x94\x893\x1a\x7f\x93\xd5\x14\x05'
sage: cipher = AES.new(key, AES.MODE_CBC, iv)
sage:
sage: encrypted = b'\xe3Nf1\xd4\xf7g\xad\xa6\xa4q\x8e\x85\x99\xa2/>\xb0u\x16\x1f\xc5\x1e\x8a\xf3\xd0t\xf5\xc4F\x9a\xce'
sage:
sage: decrypted = unpad(cipher.decrypt(encrypted), 16)
sage: print(decrypted.decode())
HTB{ORD3R_..._0RD3R_!!!}