## 1 Introduction

In 1978, Rivest, Shamir and Adleman [19] invented the RSA cryptosystem. Nowadays, it is the most widely used public key cryptosystem and serves for encryption and signature. The security of RSA is based on the difficulty of factoring specific large integers, called RSA moduli. An RSA modulus is in the form $$N=pq$$ where p and q are large prime numbers of the same size. The public exponent in RSA is an integer e satisfying $$\gcd (e,(p-1)(q-1))=1$$ while the private exponent is the integer d satisfying $$ed\equiv 1\pmod {(p-1)(q-1)}$$. Since its invention, the RSA cryptosystem has been intensively studied for vulnerabilities. Many attacks on RSA exploit the RSA key equation $$ed-k(p-1)(q-1)=1$$. A few attacks are based on the continued fraction algorithm such as Wiener’s attack [22] and most of the attacks are based on lattice reduction techniques, introduced by Coppersmith [8] (see [2, 3, 10, 15]). Combining both techniques, Blömer and May [1] presented an attack using the generalized key equation $$ex+y=k(p-1)(q-1)$$ for suitably small integers x, k and y.

Many variants of RSA have been proposed for improving the security or reducing the encryption or the decryption time (see [4, 18, 21]). The variants of RSA in [7, 9, 13, 20] make use of a public exponent e and a private exponent d satisfying the equation

\begin{aligned} ed-k\left( p^2-1\right) \left( q^2-1\right) =1. \end{aligned}
(1)

In [5], Bunder et al. proposed an attack on these variants by using the continued fraction algorithm approach. Setting $$e=N^\beta$$, they showed that one can solve the Eq. 1 and find the prime factors p and q if $$d=N^\delta$$ and $$\delta <\frac{1}{2}(3-\beta )$$. This was recently improved to $$\delta <2-\sqrt{\beta }$$ by Peng et al. [17] and by Zheng et al. [23] by using lattice reduction techniques and Coppersmith’s method.

In this paper we consider the generalized equation

\begin{aligned} eu-\left( p^2-1\right) \left( q^2-1\right) v=w. \end{aligned}
(2)

This equation can be transformed into the modular equation

\begin{aligned} v(p+q)^2-(N+1)^2v-w\equiv 0\pmod e. \end{aligned}
(3)

We set $$e=N^\beta$$, $$u=N^\delta$$, $$w=N^\gamma$$ and using lattice reduction techniques and Coppermith’s method, we show that one can solve the Eq. (3) and find the prime factors p and q under the condition

\begin{aligned} \delta < \frac{7}{3}-\gamma -\frac{2}{3}\sqrt{1+3\beta -3\gamma }-\varepsilon , \end{aligned}
(4)

where $$\varepsilon$$ is a small positive constant. Observe that the key Eq. (1) is a special case of the Eq. (3) where $$w=1$$ and $$\gamma =0$$. In this special case, the condition (4) becomes

\begin{aligned} \delta < \frac{7}{3}-\frac{2}{3}\sqrt{1+3\beta }-\varepsilon , \end{aligned}

which is slightly worst than the condition $$\delta <2-\sqrt{\beta }$$ derived by the method of Peng et al. [17]. Apart this special case, our method supersedes the method of Peng et al. since their method works only for $$w=1$$ while our method works for any $$w=N^\gamma$$ under the condition (4).

In [6], Bunder et al. studied the Eq. (2) using a combination of the continued fraction algorithm and Coppersmith’s method. They showed that this equation can be solved whenever

\begin{aligned} uv<2N-4\sqrt{2}N^{\frac{3}{4}} \quad \text {and}\quad |w|<(p-q)N^{\frac{1}{4}}v. \end{aligned}

The first condition implies the following one

$$\delta <\frac{3-\beta }{2},$$

which is worst than our condition with $$\gamma =0$$. As a consequence, our new method can be seen as an extension of the method of Bunder et al. [6].

The rest of the paper is organized as follows. In Sect. 2, we briefly describe the RSA variants that use exponents satisfying $$ed\equiv 1\pmod {\left( p^2-1\right) \left( q^2-1\right) }$$. We also recall some facts on Coppersmith’s method and lattice basis reduction. In Sect. 3, we present our attack. In Sect. 4, we present a comparison with existing attacks. We conclude the paper in Sect. 5.

## 2 Preliminaries

In this section, we briefly present some variants of the RSA cryptosystem that use the key equation $$ed\equiv 1\pmod {\left( p^2-1\right) \left( q^2-1\right) }$$. We also present Coppersmith’s method and lattice basis reduction.

### 2.1 LUC Cryptosystem

LUC cryptosystem, introduced by Smith and Lennon [20] in 1993 is based on Lucas functions. A related cryptosystem was propose by Castagnos [7] in 2007. Both cryptosystems use an RSA modulus $$N=pq$$, a public exponent e, and a private exponent satisfying a key equation $$ed-k\left( p^2-1\right) \left( q^2-1\right) =1$$ which can be generalized by the equation $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$.

### 2.2 RSA Type Schemes Based on Singular Cubic Curves

In 1995, Kuwakado, Koyama, and Tsuruoka [13] proposed a new cryptosystem based on the singular cubic with equation

$$y^2=x^3+bx^2\mod N.$$

where $$N=pq$$ is an RSA modulus. In this cryptosystem, the encryption and the decryption keys satisfy an equation of the form $$ed-k\left( p^2-1\right) \left( q^2-1\right) =1$$. A generalization of this equation is $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$.

### 2.3 RSA with Gaussian Primes

A variant of RSA was introduced in 2002 by Elkamchouchi, Elshenawy and Shaban [9]. It is an extension of the RSA cryptosystem to the domain of Guassian integers. Gaussian integers are complex numbers of the form $$z=a+bi$$ where a and b are integers and $$i^2=-1$$. The norm of a Gaussian integer is $$|a+bi|=\sqrt{a^2+b^2}$$. In the RSA variant with Gaussian integers, the modulus is $$N=PQ$$, a product of two Gaussian integer primes P and Q and the public and private exponents satisfy $$ed-k\left( |P|^2-1\right) \left( |Q|^2-1\right) =1$$. If $$P=p$$ and $$Q=q$$ are integer primes, then $$ed\,-\,k\left( p^2-1\right) \left( q^2-1\right) =1$$. This can be generalized as $$eu\,-\,\left( p^2-1\right) \left( q^2-1\right) v=w$$.

### 2.4 Coppersmith’s Method

In 1996, Coppersmith [8] proposed two methods related to finding small modular roots of univariate polynomials and small integer roots of bivariate polynomials. Since then, many techniques have been proposed for more variables (see [16]). Let

$$h(x,y,z)=\sum _{i,j,k}a_{i,j,k}x^iy^jz^k\in \mathbb {Z}[x,y,z],$$

be a polynomial with $$\omega$$ monomials. Its Euclidean norm is

$$\Vert h(x,y,z)\Vert =\sqrt{\sum _{i,j,k}a_{i,j,k}^2}.$$

The following result was proposed by Howgrave-Graham [11] to find the small modular roots of a polynomial.

### Theorem 1

Let e be a positive integer and $$h(x,y,z)\in \mathbb {Z}[x,y,z]$$ be a polynomial with at most $$\omega$$ monomials. Suppose that

\begin{aligned} \Vert h(xX,yY,zZ)\Vert <\frac{e^m}{\sqrt{\omega }} \quad \text {and}\quad h\left( x_0,y_0,z_0\right) \equiv 0\pmod {e^m}, \end{aligned}

where $$|x_0|<X$$, $$|y_0|<Y$$, $$|z_0|<Z$$. Then $$h\left( x_0,y_0,z_0\right) =0$$ holds over the integers.

Coppersmith’s method enables to find several polynomials that can be used in Howgrave-Graham’s Theorem 1. This is possible by applying a lattice reduction technique such as the LLL algorithm [14] to a lattice with a given basis. In general, the LLL algorithm produces a reduced basis with relatively small norms such as in the following result (see [15]).

### Theorem 2

(LLL). Let $$\mathcal {L}$$ be a lattice spanned by a basis $$(u_1,\ldots ,u_\omega )$$. Then the LLL algorithm outputs a new basis $$(b_1,\ldots ,b_\omega )$$ satisfying

$$\Vert b_1\Vert \le \ldots \le \Vert b_i\Vert \le 2^\frac{\omega (\omega -1)}{4(\omega +1-i)}\det (\mathcal {L})^\frac{1}{\omega +1-i},\quad i=1,\ldots ,\omega -1,$$

where $$\det (\mathcal {L})$$ is the determinant of the lattice.

We assume that if $$h_1,h_2,h_3\in \mathbb {Z}[x,y,z]$$ are three polynomials produced by Coppersmith’s method, then the ideal generated by the polynomial equations $$h_1(x,y,z) = 0$$, $$h_2(x,y,z) = 0$$, $$h_3(x,y,z) = 0$$ has dimension zero. Then, a system of polynomials sharing the root can be solved by using Gröbner basis computation or resultant techniques.

## 3 The Attack

### Theorem 3

Let $$N=pq$$ be an RSA modulus and $$e=N^\beta$$ be a public exponent. Suppose that e satisfies the equation $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$ with $$u<N^\delta$$ and $$|w|<N^\gamma$$. If

\begin{aligned} \delta < \frac{7}{3}-\gamma -\frac{2}{3}\sqrt{1+3\beta -3\gamma }-\varepsilon , \end{aligned}

then one can factor N in polynomial time.

### Proof

Let $$N=pq$$ be an RSA modulus. Let e be a public exponent satisfying $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$ with $$|w|<eu$$. Suppose that $$e=N^\beta$$, $$u<N^\delta$$ and $$|w|<N^\gamma$$. Then

\begin{aligned} v=\frac{eu-w}{\left( p^2-1\right) \left( q^2-1\right) }<\frac{eu+|w|}{\left( p^2-1\right) \left( q^2-1\right) }<2N^{\beta +\delta -2}, \end{aligned}

where we used $$\left( p^2-1\right) \left( q^2-1\right) \approx N^2$$. It follows that the solution (uvw) of the equation $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$ satisfies $$u<N^\delta$$, $$v<2N^{\beta +\delta -2}$$ and $$|w|<N^\gamma$$. We set

\begin{aligned} X=2N^{\beta +\delta -2}, Y=3N^{\frac{1}{2}}, Z=N^{\gamma }. \end{aligned}
(5)

This means that the solution (uvw) satisfies $$u<N^\delta$$, $$v<X$$ and $$|w|<Z$$. Moreover, since p and q are of the same size, then we have $$p+q<3N^{\frac{1}{2}}=Y$$.

Transforming the equation $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$, we get a modular one, namely $$-v\left( (N+1)^2-(p+q)^2\right) -w\equiv 0\pmod e.$$ This can be rewritten as

$$v(p+q)^2-(N+1)^2v-w\equiv 0\pmod e.$$

Consider the polynomial

$$f(x,y,z)=xy^2+a_1x+z,$$

where $$a_1=-(N+1)^2$$. Then $$(x,y,z)=(v,p+q,-w)$$ is a solution of the polynomial modular equation $$f(x,y,z)\equiv 0\pmod e$$. To find the small solutions of the equation $$f(x,y,z)\equiv 0\pmod {e}$$, we apply Coppersmith’s method combined with the extended strategy of Jochemsz and May [12] for finding small modular roots.

Let m and t be positive integers to be specified later. For $$0\le k\le m$$, define the set

A straightforward calculation shows that $$f^{m}(x,y,z)$$ is

\begin{aligned} f^{m}(x,y,z) =\sum _{i_1=0}^m\sum _{i_2=0}^{i_1} \left( {\begin{array}{c}m\\ i_1\end{array}}\right) \left( {\begin{array}{c}i_1\\ i_2\end{array}}\right) a_1^{i_1-i_2}x^{i_1}y^{2i_2}z^{m-i_1}. \end{aligned}

Hence, $$x^{i_1}y^{2i_2}z^{i_3}$$ is a monomial of $$f^m(x,y,z)$$ if

$$i_1 = 0,\ldots ,m,\quad i_2 = 0,\ldots ,i_1,\quad i_3=m-i_1.$$

Similarly, $$x^{i_1}y^{2i_2}z^{i_3}$$ is a monomial of $$f^{m-k}(x,y,z)$$ if

$$i_1 = 0,\ldots ,m-k,\quad i_2 = 0,\ldots ,i_1,\quad i_3=m-k-i_1.$$

From this, we deduce that for $$0\le k\le m$$, if $$x^{i_1}y^{2i_2}z^{i_3}$$ is a monomial of $$f^m(x,y,z)$$, then $$\frac{x^{i_1}y^{2i_2}z^{i_3}}{\left( xy^2\right) ^k}$$ is a monomial of $$f^{m-k}(x,y,z)$$ if

$$i_1 = k,\ldots ,m,\quad i_2 = k,\ldots ,i_1,\quad i_3 = m-i_1.$$

This leads to a characterization of the set $$M_k$$. For $$0\le k\le m$$, we obtain

$$x^{i_1}y^{i_2}z^{i_3}\in M_k\ \text {if}\ i_1 = k,\ldots ,m,\ i_2 = 2k,\ldots ,2i_1+t,\ i_3 = m-i_1.$$

Replacing k by $$k+1$$, we get

\begin{aligned} \begin{aligned}&x^{i_1}y^{i_2}z^{i_3}\in M_{k+1}\ \text {if}\\&i_1 = k+1,\ldots ,m,\ i_2 = 2k+2,\ldots ,2i_1+t,\ i_3 = m-i_1. \end{aligned} \end{aligned}

For $$0\le k\le m$$, define the polynomials

$$g_{k,i_1,i_2,i_3}(x,y,z)=\frac{x^{i_1}y^{i_2}z^{i_3}}{\left( xy^2\right) ^k}f(x,y,y)^ke^{m-k}\quad \text {with}\quad x^{i_1}y^{i_2}z^{i_3}\in M_k\big \backslash M_{k+1}.$$

Since for $$t\ge 1$$, we have

\begin{aligned} \begin{aligned}&x^{i_1}y^{i_2}z^{i_3}\in M_k\big \backslash M_{k+1}\\&\text {if}\ i_1 = k,\ldots ,m,\ i_2 = 2k,2k+1,\ i_3 = m-i_1,\\&\text {or}\ i_1 = k,\ i_2 = 2k+2,\ldots , 2i_1+t,\ i_3 = m-i_1, \end{aligned} \end{aligned}

then the polynomials $$g_{k,i_1,i_2,i_3}(x,y,z)$$ reduce to the polynomials $$G_{k,i_1,i_2,i_3}(x,y,z)$$ and $$H_{k,i_1,i_2,i_3}(x,y,z)$$ where

\begin{aligned} \begin{aligned} G_{k,i_1,i_2,i_3}(x,y,z)&=x^{i_1-k}y^{i_2-2k}z^{i_3}f(x,y,z)^ke^{m-k},\\&\text {for}\quad k=0,\ldots m,\ i_1 = k,\ldots ,m,\ i_2 = 2k,2k+1,\ i_3 = m-i_1,\\ H_{k,i_1,i_2,i_3}(x,y,z)&=y^{i_2-2k}z^{i_3}f(x,y,z)^ke^{m-k},\\&\text {for}\quad k=0,\ldots m,\ i_1 = k,\ i_2 = 2k+2,\ldots ,2i_1+t,\ i_3 = m-i_1. \end{aligned} \end{aligned}

Observe that for the target solution $$(x,y,z)=(v,p+q,-w)$$, the former polynomials satisfy

$$G_{k,i_1,i_2,i_3}(x,y,z)\equiv H_{k,i_1,i_2,i_3}(x,y,z)\equiv 0\pmod {e^m}.$$

Let $$\mathcal {L}$$ denote the lattice spanned by the coefficient vectors of the polynomials $$G_{k,i_1,i_2,i_3}(xX,yY,zZ)$$ and $$H_{k,i_1,i_2,i_3}(xX,yY,zZ)$$ where X, Y and Z are positive integers to be defined later. The ordering of rows is such that any polynomial $$G_{k,i_1,i_2,i_3}(xX,yY,zZ)$$ is prior to any polynomial $$H_{k,i_1,i_2,i_3}(xX,yY,zZ)$$. Inside each type of polynomial, the ordering of the tuples $$(k,i_1,i_2,i_3)$$ follows rule

$$(k,i_1,i_2,i_3) \prec (k',i_1',i_2',i_3') \text { if } {\left\{ \begin{array}{ll} k<k',\\ k=k', i_1<i_1'\\ k=k', i_1=i_1', i_2<i_2',\\ k=k', i_1=i_1', i_2=i_2', i_3<i_3'. \end{array}\right. }$$

Similarly, the monomials $$x^{i_1}y^{i_1}z^{i_1}$$ in the columns are ordered following the rule

$$x^{i_1}y^{i_1}z^{i_1}\prec x^{i_1'}y^{i_2'}z^{i_3'} \text { if } {\left\{ \begin{array}{ll} i_1<i_1'\\ i_1=i_1', i_2<i_2',\\ i_1=i_1', i_2=i_2', i_3<i_3'. \end{array}\right. }$$

This leads to a left triangular matrix. As an example, for $$m=2$$ and $$t=3$$, the matrix is presented in the following triangular table where the non-zero terms are denoted $$*$$.

$$\tiny { \begin{array}{c|ccccccccccccccccccc} \text {Polynomial}&{}z^2&{}yz^2&{}xz&{}xyz&{}x^2&{}x^2y&{}xy^2z&{}xy^3z&{}x^2y^2&{}x^2y^3&{}x^2y^4&{}x^2y^5&{}y^2z^2 &{}y^3z^2&{}xy^4z&{}xy^5z&{}x^2y^6&{}x^2y^7\\ G_{0, 0, 0, 2}&{}Z^2e^2&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{0, 0, 1, 2}&{}0&{}YZ^2e^2&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{0, 1, 0, 1}&{}0&{}0&{}XZe^2&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{0, 1, 1, 1}&{}0&{}0&{}0&{}XYZe^2&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{0, 2, 0, 0}&{}0&{}0&{}0&{}0&{}X^2e^2&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{0, 2, 1, 0}&{}0&{}0&{}0&{}0&{}0&{}X^2Ye^2&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{1, 1, 2, 1}&{}*&{}0&{}*&{}0&{}0&{}0&{}ZXY^2e&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{1, 1, 3, 1}&{}0&{}*&{}0&{}*&{}0&{}0&{}0&{}Y^3ZXe&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{1, 2, 2, 0}&{}0&{}0&{}*&{}0&{}*&{}0&{}0&{}0&{}X^2Y^2e&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{1, 2, 3, 0}&{}0&{}0&{}0&{}*&{}0&{}*&{}0&{}0&{}0&{}X^2Y^3e&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{2, 2, 4, 0}&{}*&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}X^2Y^4&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ G_{2, 2, 5, 0}&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}X^2Y^5&{}0&{}0&{}0&{}0&{}0&{}0\\ \hline H_{0, 0, 2, 2}&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}Y^2Z^2e^2&{}0&{}0&{}0&{}0&{}0\\ H_{0, 0, 3, 2}&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}Y^3Z^2e^2&{}0&{}0&{}0&{}0\\ H_{1, 1, 4, 1}&{}0&{}0&{}0&{}0&{}0&{}0&{}*&{}0&{}0&{}0&{}0&{}0&{}*&{}0&{}Y^4ZXe&{}0&{}0&{}0\\ H_{1, 1, 5, 1}&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}*&{}0&{}0&{}0&{}0&{}0&{}*&{}0&{}Y^5ZXe&{}0&{}0\\ H_{2, 2, 6, 0}&{}0&{}0&{}0&{}0&{}0&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}Y^6X^2&{}0\\ H_{2, 2, 7, 0}&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}*&{}0&{}Y^7X^2 \end{array}}$$

Since the matrix is triangular, then only the diagonal terms contribute to the determinant. On the other hand, only e, X, Y and Z contribute to the determinant and we get the form

\begin{aligned} \det (\mathcal {L})=e^{n_e}X^{n_X}Y^{n_Y}Z^{n_Z}. \end{aligned}
(6)

Using the construction of the polynomials $$G_{k,i_1,i_2,i_3}(x,y,z)$$ and $$H_{k,i_1,i_2,i_3}(x,y,z)$$, the exponents $$n_e$$, $$n_X$$, $$n_Y$$, $$n_Z$$, and the dimension $$\omega$$ of the lattice are as follows

\begin{aligned} \begin{aligned} n_e&=\sum _{k=0}^m\sum _{i_1=k}^m\sum _{i_2=2k}^{2k+1}\sum _{i_3=m-i_1}^{m-i_1}(m-k) +\sum _{k=0}^m\sum _{i_1=k}^k\sum _{i_2=2k+2}^{2i_1+t}\sum _{i_3=m-i_1}^{m-i_1}(m-k)\\&=\frac{1}{6}m(m+1)(4m+3t+5),\\ n_X&=\sum _{k=0}^m\sum _{i_1=k}^m\sum _{i_2=2k}^{2k+1}\sum _{i_3=m-i_1}^{m-i_1}i_1 +\sum _{k=0}^m\sum _{i_1=k}^k\sum _{i_2=2k+2}^{2i_1+t}\sum _{i_3=m-i_1}^{m-i_1}i_1\\&=\frac{1}{6}m(m+1)(4m+3t+5),\\ n_Y&=\sum _{k=0}^m\sum _{i_1=k}^m\sum _{i_2=2k}^{2k+1}\sum _{i_3=m-i_1}^{m-i_1}i_2 +\sum _{k=0}^m\sum _{i_1=k}^k\sum _{i_2=2k+2}^{2i_1+t}\sum _{i_3=m-i_1}^{m-i_1}i_2\\&=\frac{1}{6}(m+1)\left( 4m^2+6mt+3t^2+5m+3t\right) ,\\ n_Z&=\sum _{k=0}^m\sum _{i_1=k}^m\sum _{i_2=2k}^{2k+1}\sum _{i_3=m-i_1}^{m-i_1}i_3 +\sum _{k=0}^m\sum _{i_1=k}^k\sum _{i_2=2k+2}^{2i_1+t}\sum _{i_3=m-i_1}^{m-i_1}i_3\\&=\frac{1}{6}m(m+1)(2m+3t+1).\\ \omega&=\sum _{k=0}^m\sum _{i_1=k}^m\sum _{i_2=2k}^{2k+1}\sum _{i_3=m-i_1}^{m-i_1}1 +\sum _{k=0}^m\sum _{i_1=k}^k\sum _{i_2=2k+2}^{2i_1+t}\sum _{i_3=m-i_1}^{m-i_1}1\\&=(m+1)(m+t+1). \end{aligned} \end{aligned}
(7)

For $$t=\tau m$$ and sufficiently large m, we can approximate the exponents $$n_e$$, $$n_X$$, $$n_Y$$, $$n_Z$$ by their leading term and get

\begin{aligned} \begin{aligned} n_e&=\frac{1}{6}(3\tau +4)m^3+o(m^3),\\ n_X&=\frac{1}{6}(3\tau +4)m^3+o(m^3),\\ n_Y&=\frac{1}{6}(3\tau ^2+6\tau +4)m^3+o(m^3),\\ n_Z&=\frac{1}{6}(3\tau +2)m^3+o(m^3),\\ \omega&=(\tau +1)m^2+o(m^2). \end{aligned} \end{aligned}
(8)

Applying the LLL algorithm to the lattice $$\mathcal {L}$$, we get a reduced basis where the three first vectors $$h_i(Xx,Yy,Zz)$$, $$i=1,2,3$$ satisfy the conditions $$\Vert h_1(Xx,Yy,Zz)\Vert \le \Vert h_2(Xx,Yy,Zz)\Vert \le \Vert h_3(Xx,Yy,Zz)\Vert ,$$ and

$$\Vert h_3(Xx,Yy,Zz)\Vert \le 2^\frac{\omega (\omega -1)}{4(\omega -2)}\det (\mathcal {L})^\frac{1}{\omega -2}.$$

For comparison, Theorem 1 can be applied if

$$\Vert h_3(Xx,Yy,Zz)\Vert <\frac{e^m}{\sqrt{\omega }}.$$

To this end, we set

$$2^\frac{\omega (\omega -1)}{4(\omega -2)}\det (\mathcal {L})^\frac{1}{\omega -2} <\frac{e^m}{\sqrt{\omega }},$$

or equivalently

$$\det (\mathcal {L}) <\frac{2^{-\frac{\omega (\omega -1)}{4}}}{\left( \sqrt{\omega }\right) ^{\omega -2}} e^{m(\omega -2)}.$$

Hence, using (6), we get

\begin{aligned} e^{n_e-m\omega }X^{n_X}Y^{n_Y}Z^{n_Z}< \frac{2^{-\frac{\omega (\omega -1)}{4}}}{\left( \sqrt{\omega }\right) ^{\omega -2}} e^{-2m}, \end{aligned}
(9)

where the right side term is a small constant depending only on e and m. Plugging the values of $$n_e$$, $$n_X$$, $$n_Y$$, $$n_Z$$ and $$\omega$$ from (8) as well as the values $$e= N^\beta$$, $$X=2N^{\beta +\delta -2}$$, $$Y=3N^{\frac{1}{2}}$$, $$Z=N^{\gamma }$$ in each term of (9), we get

\begin{aligned} \begin{aligned} e^{n_e-m\omega }&=N^{\left( -\frac{1}{2}\tau -\frac{1}{3}\right) \beta m^3\,+\,o(m^3)},\\ X^{n_X}&=N^{\left( \frac{1}{2}\tau \,+\,\frac{2}{3}\right) (\beta \,+\,\delta -2)m^3\,+\,o(m^3)}\cdot 2^{\left( \frac{1}{2}\tau \,+\,\frac{2}{3}\right) m^3\,+\,o(m^3)}\\&=N^{\left( \frac{1}{2}\tau \,+\,\frac{2}{3}\right) (\beta \,+\,\delta -2)m^3\,+\,o(m^3)\,+\,\varepsilon _1},\\ Y^{n_Y}&=N^{\frac{1}{2}\left( \frac{1}{2}\tau ^2\,+\,\tau \,+\,\frac{2}{3}\right) m^3\,+\,o(m^3)}\cdot 3^{\left( \frac{1}{2}\tau ^2\,+\,\frac{1}{2}\tau \,+\,\frac{1}{6}\right) m^3\,+\,o(m^3)}\\&=N^{\frac{1}{2}\left( \frac{1}{2}\tau ^2\,+\,\tau \,+\,\frac{2}{3}\right) m^3\,+\,o(m^3) \,+\,\varepsilon _2},\\ Z^{n_Z}&=N^{\left( \frac{1}{2}\tau \,+\,\frac{1}{3}\right) \gamma m^3\,+\,o(m^3)},\\ \frac{2^{-\frac{\omega (\omega -1)}{4}}}{\left( \sqrt{\omega }\right) ^{\omega -2}} e^{-2m}&=N^{-2\beta m-\varepsilon _3}, \end{aligned} \end{aligned}

where $$\varepsilon _1$$, $$\varepsilon _2$$ and $$\varepsilon _3$$ are small positive constants depending on m, and N. It follows that the inequality (9) can be rewritten in terms of the exponents as

\begin{aligned} \begin{aligned} \left( -\frac{1}{2}\tau -\frac{1}{3}\right)&\beta +\left( \frac{1}{2}\tau +\frac{2}{3}\right) (\beta +\delta -2)\\&+\frac{1}{2}\left( \frac{1}{2}\tau ^2+\tau +\frac{2}{3}\right) +\left( \frac{1}{2}\tau +\frac{1}{3}\right) \gamma < \frac{-2\beta m-\varepsilon _3-\varepsilon _1-\varepsilon _2}{m^3}. \end{aligned} \end{aligned}

Setting $$\frac{-2\beta m\,-\,\varepsilon _3\,-\,\varepsilon _3\,-\,\varepsilon _1\varepsilon _2}{m^3}=-\varepsilon _4$$ and rearranging, we get

\begin{aligned} 3\tau ^2+6(\delta +\gamma -1)\tau +4\beta +8\delta +4\gamma -12<-12\varepsilon _4. \end{aligned}
(10)

The left side of (10) is optimal for $$\tau _0=1-\delta -\gamma .$$ Plugging $$\tau _0$$ in (10), we get

$$-3\delta ^2+(14-6\gamma )\delta -\gamma ^2+4\beta +10\gamma -15<-12\varepsilon _4.$$

This inequality is valid if

\begin{aligned} \delta < \frac{7}{3}-\gamma -\frac{2}{3}\sqrt{1+3\beta -3\gamma }-\varepsilon , \end{aligned}
(11)

where $$\varepsilon$$ is a small positive constant depending on m and N. This terminates the proof.   $$\square$$

## 4 Comparison with Existing Results

In [6], Bunder et al. combined the continued fraction algorithm and Coppersmith’s method to study the equation $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$. They showed that it is possible to solve it if

\begin{aligned} uv<2N-4\sqrt{2}N^{\frac{3}{4}} \quad \text {and}\quad |w|<(p-q)N^{\frac{1}{4}}v. \end{aligned}

In terms of $$e=N^\beta$$, $$u=N^\delta$$ and $$|w|=N^\gamma$$, the first condition implies the following one

$$\delta <\frac{3-\beta }{2}.$$

For $$\gamma =0$$, that is $$w=1$$, the bound of Theorem 3 becomes

\begin{aligned} \delta < \frac{7}{3}-\frac{2}{3}\sqrt{1+3\beta }-\varepsilon . \end{aligned}

Neglecting the $$\varepsilon$$ term, the difference between the former bound and the bound of [6] is

$$\delta _1=\frac{7}{3}-\frac{2}{3}\sqrt{1+3\beta }-\frac{3-\beta }{2}= \frac{5}{6}+\frac{b}{2}-\frac{2}{3}\sqrt{1+3\beta }.$$

A straightforward calculation shows that $$\delta _1\ge 0$$. This shows that the bound of Theorem 3 is better than the bound of [6].

In [17], Peng et al. proposed a lattice based method to solve the equation $$ed-k\left( p^2-1\right) \left( q^2-1\right) =1$$ under the condition $$\delta <2-\sqrt{\beta }$$ and $$\beta >1$$. This is a special case of the general equation $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$. In this special case, we have $$w=N^\gamma =1$$ and $$\gamma =0$$, and the difference between the bound of Theorem 3 and the bound of [17] is

$$\delta _2=2-\sqrt{\beta }-\left( \frac{7}{3}-\frac{2}{3}\sqrt{1+3\beta }\right) = \frac{2}{3}\sqrt{1+3\beta }-\frac{1}{3}-\sqrt{\beta }.$$

Again, a straightforward calculation shows that $$\delta _2\ge 0$$. This means that the condition of Theorem 3 is not better than Peng al.’s bound. Nevertheless, our method is more general and can solve a variety of equations with $$w\ne 1$$.

## 5 Conclusion

In this paper, we have studied the equation $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$ which is a generalization of the equation $$ed-k\left( p^2-1\right) \left( q^2-1\right) =1$$. The latter equation is the key equation of some variants of the RSA cryptosystem with modulus $$N=pq$$, public exponent e and private key d. We have showed that, under some conditions, it is possible to solve the equation $$eu-\left( p^2-1\right) \left( q^2-1\right) v=w$$ and break the cryptosystem. The attack is based on applying Coppersmith’s method to a multivariate modular equation and can be seen as an extension of former attacks on such cryptosystems.