What does it mean for $\sigma^N$≡$\sigma^e$ (mod N) for arbitrary a?

106 Views Asked by At

Full disclosure: I am trying to solve a homework problem.

As part of a homework problem, we were given a large semiprime $N$, which was used as both the modulus and the public key and asked to perform various RSA operations/attacks on it. We were then also given an $e'$, which also satisfied the requirement of a public key for verification. Given this information, we were asked to factor $N$. I've got it down to that for arbitrary $\sigma$, this relation is satisfied:

$$\sigma^N\equiv\sigma^{e'}(mod\ N)$$

I'm not seeing the significance of this relation, but I'm pretty sure it has something to do with Euler's theorem

1

There are 1 best solutions below

0
On BEST ANSWER

Write $N=pq$ for primes $p$ and $q$. Pick $\sigma$ in such a way that it's a primitive root modulo $p$ and modulo $q$. This means that the multiplicative order of $\sigma$ modulo $N$ is $\frac12(p-1)(q-1)$ and since $\gcd(N,e)=1$, it follows from $$\sigma^N\equiv \sigma^e\pmod N$$ that $$\sigma^{N-e}\equiv 1\pmod N$$ and therefore $\frac12(p-1)(q-1)\mid N-e$. For large enough $N$ this gives either $$\begin{cases}N&=pq\\N-e&=\frac12(p-1)(q-1)\end{cases}$$ or $$\begin{cases}N&=pq\\N-e&=(p-1)(q-1)\end{cases}.$$ You should be able to take it from there.