newcrypt

Description

Solution

The script generated 4 numbers by:

for i in range(4):
    x = getPrime((1<<10)-random.randint(1<<0,1<<8))
    y = inverse(x,(self.p-1)*(self.q-1)//GCD(self.p-1,self.q-1))
    numbers.append(y)

The vulnerabililty here is that xs' bit length is small compared to n, so lattice-based attack could exploit it.

Howgrave-Graham and Seifert’s Attack had concluded that when given 4 public exponents whose private exponents' bit length is below

$\delta =0.441$ times the bit length of n, it's likely to factorize n efficiently.

The bit length of x here is about

$\frac{1024-128}{2048}=0.4375$.

The math

$e_i{d}_i=1+k_i\frac{\varphi }{G}$

By this relation of e and d, we can have 10 basic equations: (

$G$ is GCD and

$s=n-\varphi$)

$W_i:G{e}_i{d}_i-k_{i}n=G-k_{i}s$

$G_{i,j}:k_{i}G{e}_j{d}_j-k_{j}G{e}_i{d}_i=G(k_i-k_j)$

The basic idea is using these equations to construct a vector-matrix equation

$xB=v$ where

$B$ is an upper triangle matrix and

$x=(k_1{k}_2{k}_3{k}_4,d_1{k}_2{k}_3{k}_4,k_1{d}_2{k}_3{k}_4,d_1{d}_2{k}_3{k}_4,k_1{k}_2{d}_3{k}_4,...,d_1{d}_2{d}_3{d}_4)$

For the purpose that will be mentioned later, we want to choose

$G_{i,j}$ as many as possible and

$W_i$ as less as possible.

My choice was:

$$\begin{gathered} k_1{k}_2{k}_3{k}_4=k_1{k}_2{k}_3{k}_4,k_2{k}_3{k}_4{W}_1,k_3{k}_4{G}_{1,2},k_3{k}_4{W}_1{W}_2, \\ {k}_2{k}_4{G}_{1,3},k_4{W}_1{G}_{2,3},k_4{W}_2{G}_{1,3},k_4{W}_1{W}_2{W}_3, \\ {k}_2{k}_3{G}_{1,4},k_3{W}_1{G}_{2,4},G_{1,2}{G}_{3,4},W_1{W}_2{G}_{3,4} \\ {G}_{1,3}{G}_{2,4},W_1{W}_3{G}_{2,4},W_2{W}_3{G}_{1,4},W_1{W}_2{W}_3{W}_4 \end{gathered}$$

This lead to:

    B=[ [1,-n,0,n^2,0,0,0,-n^3,0,0,0,0,0,0,0,n^4],
        [0,e1,-e1,-e1*n,-e1,0,e1*n,e1*n^2,-e1,0,0,0,0,0,-e1*n^2,-e1*n^3],
        [0,0,e2,-e2*n,0,e2*n,0,e2*n^2,0,e2*n,0,0,0,-e2*n^2,0,-e2*n^3],
        [0,0,0,e1*e2,0,-e1*e2,-e1*e2,-e1*e2*n,0,-e1*e2,0,0,e1*e2,e1*e2*n,e1*e2*n,e1*e2*n^2],
        [0,0,0,0,e3,-e3*n,-e3*n,e3*n^2,0,0,0,-e3*n^2,0,0,0,-e3*n^3],
        [0,0,0,0,0,e1*e3,0,-e1*e3*n,0,0,e1*e3,e1*e3*n,0,0,e1*e3*n,e1*e3*n^2],
        [0,0,0,0,0,0,e2*e3,-e2*e3*n,0,0,-e2*e3,e2*e3*n,-e2*e3,e2*e3*n,0,e2*e3*n^2],
        [0,0,0,0,0,0,0,e1*e2*e3,0,0,0,-e1*e2*e3,0,-e1*e2*e3,-e1*e2*e3,-e1*e2*e3*n],
        [0,0,0,0,0,0,0,0,e4,-e4*n,0,e4*n^2,0,e4*n^2,e4*n^2,-e4*n^3],
        [0,0,0,0,0,0,0,0,0,e1*e4,-e1*e4,-e1*e4*n,-e1*e4,-e1*e4*n,0,e1*e4*n^2],
        [0,0,0,0,0,0,0,0,0,0,e2*e4,-e2*e4*n,0,0,-e2*e4*n,e2*e4*n^2],
        [0,0,0,0,0,0,0,0,0,0,0,e1*e2*e4,0,0,0,-e1*e2*e4*n],
        [0,0,0,0,0,0,0,0,0,0,0,0,e3*e4,-e3*e4*n,-e3*e4*n,e3*e4*n^2],
        [0,0,0,0,0,0,0,0,0,0,0,0,0,e1*e3*e4,0,-e1*e3*e4*n],
        [0,0,0,0,0,0,0,0,0,0,0,0,0,0,e2*e3*e4,-e2*e3*e4*n],
        [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,e1*e2*e3*e4] ]

and

$$\begin{gathered} v=[k_1{k}_2{k}_3{k}_4,(1-k_{1}s)k_2{k}_3{k}_4,(k_1-k_2)k_3{k}_4,(1-k_{1}s)(1-k_{2}s)k_3{k}_4, \\ (k_1-k_3)k_2{k}_4,(1-k_{1}s)(k_2-k_3)k_4,(k_1-k_3)(1-k_{2}s)k_4,{\mathrm{\Pi }}_{i=1}^3(1-k_{i}s)k_4, \\ (k_1-k_4)k_2{k}_3,(1-k_{1}s)(k_2-k_4)k_3,(k_1-k_2)(k_3-k_4),(1-k_{1}s)(1-k_{2}s)(k_3-k_4), \\ (k_1-k_3)(k_2-k_4),(1-k_{1}s)(1-k_{3}s)(k_2-k_4),(1-k_{2}s)(1-k_{3}s)(k_1-k_4),{\mathrm{\Pi }}_{i=1}^4(1-k_{i}s)] \end{gathered}$$

Every row of

$B$ is a vector much bigger than

$v$ and

$v$ is linear combination of these rows.

To apply lattice attack, we still need

$v$ to be balanced. Notice that

$e_i\approx n$,

$d_i\approx n^{\delta }$,

$k_i\approx n^{\delta }$,

$s\approx n^{0.5}$, so

$v$ is dominated by the last component

$\mathrm{\Pi }(1-k_{i}s)\approx n^{2+4\delta }$.

Hence we need to multiply both

$B$ and

$v$ by a diagonal matrix

$D$

diagonal_matrix([n^2,n^1.5,n^(2+delta),n,n^(2+delta),
                 n^(1.5+delta),n^(1.5+delta),n^0.5,
                 n^(2+delta),n^(1.5+delta),n^(2+2*delta),n^(1+delta),
                 n^(2+2*delta),n^(1+delta),n^(1+delta),1])

Now we can apply LLL algorithm to obtain

$v$ and

$x$, then calculate

$\varphi =e_1\times \frac{d_1{k}_2{k}_3{k}_4}{k_1{k}_2{k}_3{k}_4}$. If

$\varphi$ is calculated correctly, the roots of

$x^2-(n+1-\varphi )x+n=0$ will be p and q consequently.

The only thing left is that we don't know the GCD of

$p-1$ and

$q-1$, it could be any even number. Hence for every possible

$G$, we need to multiply

$e$ by

$G$ first. It turned out that the GCD is just

$2$ though.

Reason why not choose

$W_i$ but

$G_{i,j}$

Using the theorem, to increase the success rate of the attack, we need the volume of

$B$ to be as large as possible. It means large components of

$D$, which means small right hand side of each equation we choose.

By some observation, all the right hand side of the equations consists of

$k_1,k_2,k_3,k_4$.

When choosing

$G_{i,j}$, we elimintae

$k_i$ and

$k_j$ but only multiply the number by

$(k_i-k_j)$. On the other hand, choosing

$W_i$ multiplies the number by

$k_{i}s$, which is much larger.

We can even calculate the upperbound of bit length of

$d_i$ by the theorem.

$$\begin{gathered} ||v||\approx n^{2+4\delta }\le n^{\frac{1}{16}(32+13\delta +22.5)}\approx 4(e_1^8{e}_2^8{e}_3^8{e}_4^8{n}^{13\delta +22.5}{)}^{\frac{1}{16}}=\sqrt{n}vol(B{)}^{\frac{1}{n}} \\ \Rightarrow \delta \le \frac{15}{34}=0.441... \end{gathered}$$

The script