Intro to RSA

Learn the basics of a common public key cryptosystem


5 solved
March 11, 2017, 10:38 p.m.


10 solved
Aug. 9, 2017, 4:37 p.m.

\phi(N) doesn't make sense here. \phi(p,q), if you want to make it a function. \phi(p,q) = (p-1)(q-1) = pq - p - q + 1 =: f(pg) - g(p,q) + 1 f(x) = x g(p,q) := p + q

There is no way of writing g(p,q) as a function h(pq). It's only possible (and indeed, useful) if the taylor series of \phi(p,q) was something a sum of (pq)^n only. Sp, not like p^3q^2 or, in our case, p^1q^0. You can prove this by contradiction if you assume it is possible and know that the function is holomorphic on its natural range (Extend to real numbers for the proof).

</autism> love the site.


10 solved
Aug. 9, 2017, 5:40 p.m.

Never mind, I'm retarded.