# Tag Info

1

No, it wouldn't. Very simply put: If you can compute square roots wrt. a composite modulus, then you can calculate other fixed roots, too. This is based on the fact, that being able to compute square roots allows you to factor the modulus: Choose random $x$. Compute $x^2$. apply your square root algorithm to $x^2$. The result $y$ will be a different ...

1

OBVIOUS, if you write $f(x)=(1+2^8+2^{16}+2^{24})\times x=K\times x$. Then $f(x)\times f(y)=K^2\times x \times y$. Then $\forall y$, choosing $z \in (\mathbb{Z}/n.\mathbb{Z})^{*}$ allows to dermine the unique $w \in \mathbb{Z}/n.\mathbb{Z}$ satisfying this relation in the multiplicative group.

1

The fact that $g$ is a generator (or not) of the group of inverse elements $G={\bf F}_p^{*}$, indeed does not affect the relation you wrote. But, if you want to apply Diffie-Hellman in a secure way, the order of $g$ has to be large. Say you choose a large prime $p$ (at least $1024$ bits). If $g$ is not a generator of $G$ then the order of $g$ shall divide ...

Top 50 recent answers are included