138 reputation
4
bio website
location Florence, Italy
age 24
visits member for 2 years, 8 months
seen Aug 15 at 15:14

Master's student in Florence, Italy, currently on an ERASMUS program at FU Berlin. My main interest is Algebraic Topology. I wrote my bachelor degree project on Morse Theory. Here are some of my favourite quotes:

  • "To ask the right question is harder than to answer it." - Georg Cantor

  • "Reductio ad absurdum, which Euclid loved so much, is one of a mathematician's finest weapons. It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game." (G. H. Hardy)

  • "We say that a proof is beautiful when it gives away the secret of the theorem, when it leads us to perceive the inevitability of the statement being proved." (Giancarlo Rota)

  • "A mathematician is a device for turning coffee into theorems." (Alfréd Rényi)

  • "Mathematicians may turn coffee into theorems, but with an american coffee the most you'll get out of me is a lemma." (a Professor at my university)

  • A topologist is someone who can't tell his ass from a hole in the ground, but can tell his ass from two holes in the ground.


Jan
5
comment Cycle attack on RSA
@PeterTaylor Yes, by " $k$ might have not much to do with the order of $\mathbb{Z}_{\phi(n)}^{\times}$" I meant in terms of magnitude - it could be a divisor and still be small.
Jan
5
comment Cycle attack on RSA
Thank you for your answer, this is what I was looking for. I still don't understand one thing though: why is the probability of $|e|$ not being a multiple of $r$ at most $1/r$?
Jan
5
comment Cycle attack on RSA
By $<e>$ i mean the group generated by $e$, $<e>=\{1,e,e^2,...,e^{k-1} \}$ (yes, $|e|$=k). I don't see the connection with factoring or with evaluating $\phi(n)$ (which are computationally equivalent, up to polynomial transformations), since in theory $k$ could be much smaller than $|\mathbb{Z}_{\phi(n)}^{\times}|=\phi(\phi(n))$ (I'm guessing it probably isn't - this is the type of result I was looking for).
Jan
5
comment Cycle attack on RSA
Then regarding the first highlighted part you quoted: I am not wondering about the order of $\space$ $\mathbb{Z}_{\phi(n)}^{\times}$ (which of course is fixed), but about the order of $e$ in $\mathbb{Z}_{\phi(n)}^{\times}$, that is the order of the subgroup $<e>$ of $\mathbb{Z}_{\phi(n)}^{\times}$. This might not have much to do with the order of $\mathbb{Z}_{\phi(n)}^{\times}$ : for example the group $\mathbb{Z}_{8}^{\times}$ has order 4, but the only possible orders of elements are $1$ or $2$, since it is the klein group.
Jan
5
comment Cycle attack on RSA
Thank you for your answer. I was expecting th first objection you made: even if I came across $m$ I might not be able to tell it's the plaintext, and distinguish it from any other element in $\mathbb{Z_n}$. I don't know much about padding, and I can see how this is possible.