# Coppersmith's Attack

Coppersmith's attack describes a class of attacks on the public-key cryptosystem RSA based on Coppersmith's theorem (see below). The public key in the RSA system is a tuple of integers $(N,e)$, where N is the product of two primes p and q. The secret key is given by an integer d satisfying $ed\equiv 1{\bmod \ }(p-1)(q-1)$; equivalently, the secret key may be given by $d_{p}\equiv d{\bmod (}p-1)$ and $d_{q}\equiv d{\bmod (}q-1)$ if the Chinese remainder theorem is used to improve the speed of decryption, see CRT-RSA. Encryption of a message M produces the ciphertext $C\equiv M^{e}{\bmod N}$ which can be decrypted using $d$ by computing $C^{d}\equiv M{\bmod N}$.

Coppersmith's theorem has many applications in attacking RSA specifically if the public exponent e is small or if partial knowledge of the secret key is available.

## Low Public Exponent Attack

In order to reduce encryption or signature-verification time, it is useful to use a small public exponent ($e$). In practice, common choices for $e$ are 3, 17 and 65537 $(2^{{16}}+1)$.[1] These values for e are Fermat primes, sometimes referred to as $F_{0},F_{2}$ and $F_{4}$ respectively $(F_{x}=2^{{2^{x}}}+1)$. They are chosen because they make the modular exponentiation operation faster. Also, having chosen such $e$, it is simpler to test whether $\gcd(e,p-1)=1$ and $\gcd(e,q-1)=1$ while generating and testing the primes in step 1 of the key generation. Values of $p$ or $q$ that fail this test can be rejected there and then. (Even better: if e is prime and greater than 2 then the test $p\,{\bmod \,}e\neq 1$ can replace the more expensive test $\gcd(p-1,e)=1$.
If the public exponent is small and the plaintext $m$ is very short, then the RSA function may be easy to invert which makes certain attacks possible. Padding schemes ensure that messages have full lengths but additionally choosing public exponent $e=2^{{16}}+1$ is recommended. When this value is used, signature-verification requires 17 multiplications, as opposed to about 25 when a random $e$ of similar size is used. Unlike low private exponent (see Wiener's Attack), attacks that apply when a small $e$ is used are far from a total break which would recover the secret key d. The most powerful attacks on low public exponent RSA are based on the following theorem which is due to Don Coppersmith.

## Theorem 1 (Coppersmith)[2]

Let N be an integer and $f\in {{\mathbb Z}}[x]$ be a monic polynomial of degree $d$ over the integers. Set $X=N^{{{\frac {1}{d}}-\epsilon }}$ for ${\frac {1}{d}}>\epsilon >0$. Then, given $\left\langle N,f\right\rangle$ attacker, Eve, can efficiently find all integers $x_{0} satisfying $f(x_{0})=0\,{\bmod \,}N$. The running time is dominated by the time it takes to run the LLL algorithm on a lattice of dimension O$(w)$ with $w={{\rm {min}}}({\frac {1}{\epsilon }},\log _{2}N)$.

This theorem states the existence of an algorithm which can efficiently find all roots of $f$ modulo $N$ that are smaller than $X=N^{{{\frac {1}{d}}}}$. As $X$ gets smaller, the algorithm's runtime will decrease. This theorem's strength is the ability to find all small roots of polynomials modulo a composite $N$.

The simplest form of Håstad's attack is presented to ease understanding. The general case uses Coppersmith's theorem.

### How does it work?[3]

Suppose one sender sends the same message $M$ in encrypted form to a number of people $P_{1};P_{2};\dots ;P_{k}$, each using the same small public exponent $e$, say $e=3$, and different moduli $\left\langle N_{i},e_{i}\right\rangle$. A simple argument shows that as soon as $k\geq 3$ ciphertexts are known, the message $M$ is no longer secure: Suppose Eve intercepts $C_{1},C_{2}$, and $C_{3}$, where $C_{i}\equiv M^{3}\,{\bmod \,}N_{i}$. We may assume $\gcd(N_{i},N_{j})=1$ for all $i,j$ (otherwise, it is possible to compute a factor of one of the $N_{i}$’s by computing $\gcd(N_{i},N_{j})$.) By the Chinese Remainder Theorem, she may compute $C\in {\mathbb {Z}}_{{N_{1}N_{2}N_{3}}}^{*}$ such that $C\equiv C_{i}\,{\bmod \,}N_{i}$. Then $C\equiv M^{3}\,{\bmod \,}N_{1}N_{2}N_{3}$ ; however, since $M for all $i$', we have $M^{3}. Thus $C=M^{3}$ holds over the integers, and Eve can compute the cube root of $C$ to obtain $M$.

For larger values of $e$ more ciphertexts are needed, particularly, $e$ ciphertexts are sufficient.

### Generalizations

Håstad also showed that applying a linear-padding to $M$ prior to encryption does not protect against this attack. Assume the attacker learns that $C_{i}=f_{i}(M)^{{e}}$ for $1\leq i\leq k$ and some linear function $f_{i}$, i.e., Bob applies a pad to the message $M$ prior to encrypting it so that the recipients receive slightly different messages. For instance, if $M$ is $m$ bits long, Bob might encrypt $M_{i}=i2^{m}+M$ and send this to the i-th recipient.

If a large enough group of people is involved, the attacker can recover the plaintext $M_{i}$ from all the ciphertext with similar methods. In more generality, Håstad proved that a system of univariate equations modulo relatively prime composites, such as applying any fixed polynomial $g_{1}(M)=0$ mod $N_{i}$, could be solved if sufficiently many equations are provided. This attack suggests that randomized padding should be used in RSA encryption.

Suppose $N_{1},\dots ,N_{k}$ are relatively prime integers and set $N_{{{\rm {min}}}}={{\rm {min}}}_{i}\{N_{i}\}$. Let $g_{i}(x)\in {\mathbb {Z}}/N_{i}\left[x\right]$ be k polynomials of maximum degree $q$. Suppose there exists a unique $M satisfying $g_{i}(M)=0$(mod $N_{i}$) for all $i\in \left\{1,\dots ,k\right\}$. Furthermore suppose $k>q$. There is an efficient algorithm which, given $\left\langle N_{i},g_{i}\left(x\right)\right\rangle$ for all $i$, computes $M$.

Proof:
Since the $N_{i}$ are relatively prime the Chinese Remainder Theorem might be used to compute coefficients $T_{i}$ satisfying $T_{i}\equiv 1{\bmod N}_{i}(is_{1})$ and $T_{i}\equiv 0{\bmod \ }N_{j}$ for all $i\neq j$. Setting $g(x)=\sum i\cdot T_{i}\cdot g_{i}(x)$ we know that $g(M)\equiv 0{\bmod \ }\prod N_{i}$. Since the $T_{i}$ are nonzero we have that $g\left(x\right)$ is also nonzero. The degree of $g\left(x\right)$ is at most $q$. By Coppersmith’s Theorem, we may compute all integer roots $x_{0}$ satisfying $g(x_{0})\equiv 0{\bmod \prod }N_{i}$ and $\left|x_{0}\right|<\left(\prod N_{i}\right)^{{{\frac {1}{q}}}}$. However, we know that $M, so $M$ is among the roots found by Coppersmith's theorem.

This theorem can be applied to the problem of broadcast RSA in the following manner: Suppose the i-th plaintext is padded with a polynomial $f_{i}\left(x\right)$, so that $N_{i}=\left(f_{i}\left(x\right)\right)^{{e_{i}}}-C_{i}{\bmod \ }N_{i}$. Then the polynomials $g_{i}=\left(f_{i}\left(x\right)\right)^{{e_{i}}}-C_{i}{\bmod N}_{i}$ satisfy that relation. The attack succeeds once $k>{{\rm {max}}}_{i}(e_{i}\cdot \deg f_{i})$. The original result used the Håstad method instead of the full Coppersmith method. Its result was required $k=O(q^{{2}})$ messages, where $q={{\rm {max}}}_{i}(e_{i}.\deg f_{i})$.[3]

## Franklin-Reiter Related Message Attack

Franklin-Reiter identified a new attack against RSA with public exponent $e=3$. If two messages differ only by a known fixed difference between the two messages and are RSA encrypted under the same RSA modulus $N$, then it is possible to recover both of them.

### How does it work?

Let $\left\langle N;e_{i}\right\rangle$ be Alice's public key. Suppose $M_{1};M_{2}\in {\mathbb {Z}}_{N}$ are two distinct messages satisfying $M_{1}\equiv f(M_{2})\,{\bmod \,}N$ for some publicly known polynomial $f\in {\mathbb {Z}}_{N}[x]$. To send $M_{1}$ and $M_{2}$ to Alice, Bob may naively encrypt the messages and transmit the resulting ciphertexts $C_{1};C_{2}$. Eve can easily recover $M_{1};M_{2}$ given $C_{1};C_{2}$, by using the following theorem:

### Theorem 3 (Franklin-Reiter)

Set $e=3$ and let $\left\langle N,e\right\rangle$ be an RSA public key. Let $M_{1}\neq M_{2}\in {\mathbb {Z}}_{N}^{*}$ satisfy $M_{1}\equiv f(M_{2})\,{\bmod \,}N$ for some linear polynomial $f=ax+b\in {\mathbb {Z}}_{N}[x]$ with $b\neq 0$. Then, given $\left\langle N,e,C_{1},C_{2},f\right\rangle$, attacker, Eve, can recover $M_{1},M_{2}$ in time quadratic in $\log _{2}N$.

For an arbitrary $e$ (rather than restricting to $e=3$) the time required is quadratic in $e$ and $\log _{2}N$).

Proof:
Since $C_{1}=M_{1}^{e}\,{\bmod \,}N$, we know that $M_{2}$ is a root of the polynomial $g_{1}(x)=f(x)^{e}-C_{1}\in {\mathbb {Z}}_{N}[x]$. Similarly, $M_{2}$ is a root of $g_{2}(x)=x^{e}-C_{2}\in {\mathbb {Z}}_{N}[x]$. The linear factor $x-M_{2}$ divides both polynomials. Therefore, Eve calculates the greatest common divisor (gcd) of $g_{1}$ and $g_{2}$, if the gcd turns out to be linear, $M_{2}$ is found. The gcd can be computed in quadratic time in $e$ and $\log _{2}N$ using the Euclidean algorithm.

Like Håstad’s and Franklin-Reiter’s attack, this attack exploits a weakness of RSA with public exponent $e=3$. Coppersmith showed that if randomized padding suggested by Håstad is used improperly then RSA encryption is not secure.

### How does it work?

Suppose Bob sends a message $M$ to Alice using a small random padding before encrypting it. An attacker, Eve, intercepts the ciphertext and prevents it from reaching its destination. Bob decides to resend $M$ to Alice because Alice did not respond to his message. He randomly pads $M$ again and transmits the resulting ciphertext. Eve now has two ciphertexts corresponding to two encryptions of the same message using two different random pads.

Even though Eve does not know the random pad being used, she still can recover the message $M$ by using the following theorem, if the random padding is too short.

### Theorem 4 (Coppersmith)

Let $\left\langle N,e\right\rangle$ be a public RSA key where $N$ is $n$-bits long. Set $m=\lfloor {\frac {n}{e^{2}}}\rfloor$. Let $M\in {\mathbb {Z}}_{N}^{*}$ be a message of length at most $n-m$ bits. Define $M_{1}=2^{m}M+r_{1}$ and $M_{2}=2^{m}M+r_{2}$, where $r_{1}$ and $r_{2}$ are distinct integers with $0\leq r_{1},r_{2}<2^{m}$. If Eve is given $\left\langle N,e\right\rangle$ and the encryptions $C_{1},C_{2}$ of $M_{1},M_{2}$ (but is not given $r_{1}$ or $r_{2}$, she can efficiently recover $M$.

Proof[2]
Define $g_{1}(x,y)=x^{e}-C_{1}$ and $g_{2}(x,y)=x^{e}-C_{2}$. We know that when $y=r_{2}-r_{1}$, these polynomials have $x=M_{1}$ as a common root. In other words, $\vartriangle =r_{2}-r_{1}$ is a root of the resultant $h(y)={{\rm {res}}}_{x}(g_{1},g_{2})\in {\mathbb {Z}}_{N}[y]$. Furthermore, $\left|\vartriangle \right|<2^{m}. Hence, $\vartriangle$ is a small root of $h$ modulo $N$, and Eve can efficiently find it using Theorem 1 (Coppersmith). Once $\vartriangle$ is known, the Franklin-Reiter attack can be used to recover $M_{2}$ and consequently $M$.