RSA encryption using multiplication

Question

Generally in RSA we encrypt as $m^e \pmod n$. Will RSA work if we replace the power by normal multiplication? $E = (m \times e) \mod n$ and decryption as $c \times d \mod n$. What will be $d$ disadvantage if it works ?

Answer 1

Suppose an RSA public key is still $(e, N)$ and the private key is still $d$. We have $Enc = (m*e) \text{ mod } N$ and $Dec = (c*d) \text{ mod } N$. The correctness of the scheme depends on the fact that $Dec(Enc(m)) = (m*e*d) \text{ mod } N = m \text{ mod } N$. This implies $e*d = 1 \text{ mod } N$. It is thus trivial to compute the private key given only the public key (i.e. compute $d = e^{-1} \text{ mod } N$). Hence this scheme is completely broken.

Note that this scheme only works if $e$ and $N$ are coprime otherwise no $d$ such that $e*d = 1 \text{ mod } N$ exists.

Answer 2

Assumption: $e$ and $d$ are prime with $N$.

If we use the multiplication instead of the power. We have $d \equiv e^{-1} \pmod N$ :

\begin{align}c \times d &= m &\mod N\\ m \times e \times d &= m &\mod N\\ e \times d &= 1 &\mod N \end{align} (It has to work for any $m$ so we can suppose $m$ prime).

$e$ and $N$ are publicly known, to compute $d$ the inverse of $e$, you just need the Extended Euclidean Algorithm : $e$ is prime with $N$, therefore there exists $u, v$ such as $$e \times u + N \times v = 1$$ by Bezout's Theorem $$e \times u = 1 \pmod N$$ Therefore $u = d = e^{-1}$ which is easily computed.