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If your ints are unsigned then the code r = (r * 33) + (int)c and the fact that you're using 32-bit integers yield the equation $\;\;\;\; \text{new_r} \: \equiv \: (\text{old_r} \cdot 33) + \text{(int)}\hspace{.02 in}\text{c} \;\; \pmod{2^{32}} \;\;\;\;$. Since 33 is odd and $2^{32}$ is even, 33 is a unit mod $2^{32}$. $\:$ I used wolframalpha to determine ...


2

Am I on the right track with reversing DJB2 (can it be reversed?)? Is there some way of finding the remainder of a large number that has been modded by 232? You were on a right track to explain why it can't be easily inverted. Given an arbitrary $h_i$, every letter of the alphabet will give you another potential $h_{i-1}$ that the value was before that ...


6

A possible RSA variant uses: some odd exponent $e>2$ (that can be $e=3$ or $e=2^{16}+1$ as customary in standard RSA); $p$ and $q$ distinct large random primes, with $\gcd(e,p)=\gcd(e,p-1)=\gcd(e,q-1)=1$; $N=p^2\cdot q$; some $d$ computed such that $d\cdot e\equiv 1\pmod{\operatorname{lcm}(p,p-1,q-1)}$; public-key function $x\to x^e\bmod N$; private-key ...


9

Actually, that was proposed here back in 1998 (sorry, an electronic version of the paper does not appear to be on the web) -- the author claimed a modest speedup in the private operations. However, that speed up would appear to be about the same if you just did "multiprime RSA", that is, selected an RSA modulus of the form $pqr$ for three distinct primes ...


1

It would probably work: RSA with composite numbers But it would be better to just choose 2 larger primes instead of reusing the same prime twice.


1

I do not know why the OpenSSL implementation specifically does this. However, a branch-less (constant time) implementation of the RSA private key operation, might be slightly more efficient if the parameter $c = q^{-1} \bmod p$ is calculated for $p$ being the greatest prime of the two. Otherwise the value of $J_q = I^{d \bmod q-1} \bmod q$ has to be taken ...


8

There's no real difference between $p$ and $q$ in RSA. It looks like OpenSSL just has the agreement "$p$ has to be bigger than $q$" for conveniences. One of the numbers has to be bigger than the other (otherwise they would be the same number, and $p = q$ is very bad in RSA). Just use two examples: $p = 13$ and $q = 11$. $p$ is bigger than $q$, all right. ...



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