# low-exponent RSA

Suppose we have 2 messages $m_1$ and $m_2$ related by a known relation $m_2=m_1+1$. Suppose further the messages are encrypted under RSA with an exponent of $3$ using a single public modulus $N$. $c_i=m_i^3$ mod$N$, $i=1,2$

Then from $c_1,c_2,N$ we can calculate the secrete messages $m_i$ algebraically as follows: $\frac{c_2+2c_1-1}{c_2-c_1+2}=\frac{3m_1^3+3m_1^2+3m_1}{3m_1^2+3m_1+3}=m_1$ mod$N$.

When the exponent is 5, setting $c_1=m^5$ mod$N$ and $c_2=(m+1)^5$ mod$N$, we can find some $P(m)$ s.t. $\frac{mP(m)}{P(m)}$. (there is explicit $P(m)$ in the above paper.)

For an arbitrary exponent $e$, the author says it will always be possible to write down an equation analogous to them.

My question is: How can I find such equations? Even, I don't know the exponent 3 case. Please help!

• You say: "I don't know the exponent 3 case". What's your issue with that? For $e=3$, given numerical values of $N$, $c_1$ and $c_2$, do you understand how to compute $m_1$ from $\frac{c_2+2c_1-1}{c_2-c_1+2}=m_1\pmod N$? Or do you "only" have difficulty deriving that equation?
– fgrieu
Jun 18 '13 at 5:16
• I can not derive that equation...Can you help me?
– Kojo
Jun 18 '13 at 8:29

Following the notation in Low-Exponent RSA with Related Messages and with $\alpha=\beta=1$ for simplification, we'll assume $e$, $N$, $c_1=m^e\bmod N$ and $c_2=(m+1)^e\bmod N$ are known; we are trying to derive $m$, in range $[0…N-1]$.

For $e=3$, a little inspiration is enough. We get

$c_2-c_1≡(m+1)^3-m^3≡3⋅m^2+3⋅m+1\pmod N$ thus
$c_2-c_1+2≡3⋅(m^2+m+1)\pmod N$

$c_2+2⋅c_1≡(m+1)^3+2⋅m^3≡3⋅m^3+3⋅m^2+3⋅m+1\pmod N$ thus
$c_2+2⋅c_1-1≡3⋅m⋅(m^2+m+1)\pmod N$

Then $m⋅(c_2-c_1+2)≡c_2+2c_1-1\pmod N$.

From which it follows that $m=(c_2-c_1+2)^{-1}⋅(c_2+2⋅c_1-1)\bmod N$ when $\gcd(c_2-c_1+2,N)=1$ (which is likely).

Updated: after fixing serious mistakes, and restructuring, I hope the following explicitly constructs an expression $m=P(c_1,c_2)^{-1}⋅Q(c_1,c_2)\bmod N$ using $\mathcal{O}(e^2)$ terms, with the expressions $P$ and $Q$ dependent on the value of $e$ but independent of the value of $N$, $c_1$, $c_2$, and allowing to compute $m$ explicitly with high odds.

Our $P(c_1,c_2)$ and $Q(c_1,c_2)$ will be of the form $$P(c_1,c_2)=\sum_{i=0}^f\sum_{j=0}^{f-i}p_{i,j}⋅c_1^i⋅c_2^j$$ $$Q(c_1,c_2)=\sum_{i=0}^f\sum_{j=0}^{f-i}q_{i,j}⋅c_1^i⋅c_2^j$$ with integer coefficients $p_{i,j}$, $q_{i,j}$, the indexes $i$, $j$ in $[0…f]$ for some $f$ to be determined, and such that $m⋅P(m^e,(m+1)^e)-Q(m^e,(m+1)^e)$ is the null polynomial in $m$.

We can develop $m⋅P(m^e,(m+1)^e)-Q(m^e,(m+1)^e)$ into a polynomial in $m$ of degree $e⋅f+1$, with each of its $e⋅f+2$ coefficients a linear combination (dependent only on the value of $e$ and $f$) of the $p_{i,j}$ and $q_{i,j}$. Asserting that this is the null polynomial is equivalent to a system of $e⋅f+2$ linear equations in $\mathbb Z$ for the $(f+1)⋅(f+2)$ unknowns $p_{i,j}$ and $q_{i,j}$. This system has a trivial solution: $p_{i,j}=q_{i,j}=0$.

When $f=e-2$ and $e\ge3$, our linear system of equations in $\mathbb Z$ has more unknowns than equations (precisely: $e-2$ more). It thus has solutions beyond the trivial one. We can find such non-trivial solution by mere linear algebra (further, we can fix any $e-3$ for our unknowns to $0$). That gives us the desired expressions for $P(c_1,c_2)$ and $Q(c_1,c_2)$. Note that the solution can't have all the $p_{i,j}=0$ (as that would imply all the $q_{i,j}=0$); thus $\gcd(P(c_1,c_2),N)=1$ is likely.