Talk:Rabin cryptosystem

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[edit] Security

The security stuff on this page is definitely wrong -- appropriate padding makes Rabin stronger, not less strong. I'll try and fix it when I have time ciphergoth 10:26, 2004 Nov 16 (UTC)

It's been fixed, I think. —Lowellian (reply) 01:50, 15 March 2006 (UTC)

[edit] Duplicate

It appears that this article is duplicated on the Rabin-Williams encryption page. I suggest that the best parts of each of the explanations be taken to form a single article and a redirect be made from the redundant page. Which page becomes the redundant page depends on which name is most appropriate. Does anyone know how this scheme came to be known under two names?

[edit] Worked Example

In the example given on the page, mq = 9. By my calculations, mq = 2.

mq2 (congruent to) c mod 11
mq2 (congruent to) 15 mod 11
mq2 = 4
mq = 2

Apologies for poor formatting.

--Newheroicideal 15:07, 25 March 2007 (UTC)

No-one seems to object, so I changed it myself. --Newheroicideal 15:18, 2 April 2007 (UTC)

In the article square roots are computed using m_q\equiv c^{(q+1)/4}\pmod{q}. Thus m_q\equiv 15^3\equiv 9\pmod{11} is the correct result. But, notice that 2\equiv -9\pmod{11}. Hence 2 is the other square root of c modulo 11, just not the one computed with described method. 85.2.2.33 20:33, 2 April 2007 (UTC)

[edit] Legendre Symbol

From c\equiv m^2\pmod{pq} follows that c is a quadratic residue. Hence \left({c\over p}\right)=1

Why is this so? Doesn't seem to be correct to me after reading an article on Jacobi symbol... —Ram3ai (talk) 21:25, 17 January 2008 (UTC)

If c\equiv m^2\pmod{pq} then c = m2 + npq for some integer n. So c\equiv m^2\pmod{p}. reetep (talk) 11:27, 18 January 2008 (UTC)