It's not that proving the Riemann Hypothesis would itself lead to a breakthrough against RSA. Rather, it's speculation that the methods leading to the discovery of a proof of the Riemann Hypothesis ...

It would take you a while, but yes. You'd have to print out several tables that calculate things for you like $GF(256)$ field multiplication and inversion, but you could do it. It would be slow and ...

When $\gcd(e, \phi(n)) = 1$, integers modulo $n$ coprime to $n$ have a unique $e$th root modulo $n$. This is the basis of RSA. Unlike for an unfactored RSA modulus, $\phi(2^{160})$ is easy to ...

Regarding AES: AES in a straightforward, single-block mode is generally written in a way that can be vulnerable to cache timing attacks, and potentially related attacks like Spectre. This is because ...

You don't have to use XOR, but rather, it tends to be convenient. One of its convenient properties is that it is its own inverse. Also, XOR implements addition in $\mathbb F_{2^n}$, making XOR a key ...

Normally, yes, the hash algorithm in use is communicated beforehand. For example, sending an algorithm identifier during the TLS/SSL handshake process. However, depending upon the "padding scheme" ...

I don't know anything about IEEE 802.11i, so I can't be sure about my answer, but hopefully this provides some insight. From the way in which the formula is written, $H$ is probably some sort of ...

After Daniel S's answer above, I wrote code to exhaustively search for all elements whose order $\le {2*23*32985101}$--the weak keys--while matching Poly1305's $r$ mask. Here is the complete list of ...

In RSA, there are various numbers that are (kind of) equivalent to the private key, but aren't the private key per se. These are numbers that if you know them, you can calculate the rest of the ...

The problem for choosing $k$ bits from $64$ ultimately comes down to computing a uniformly random integer $r$ with $0 \leq r < \frac{64!}{k!(64-k)!}$ then decoding it to determine which bits. The $... View answer 1 votes$00000001$is its own inverse in the Rijndael field, because polynomial multiplication by itself gets$00000001$and is unchanged by the modulo operation. This will always be true in any ring, as ... View answer 1 votes By definition, there exists such a function. Your function$g(z)$could be simply a table of all the possible values of$f(f(f(x, y), y), y)$mapped back to one of the possible values of$x$. I ... View answer 1 votes GMAC, for example, is trivially broken if used as an unkeyed hash algorithm. GMAC is effectively a series of operations on blocks where you take the previous state, XOR it with the next block, then ... View answer 1 votes With public-key algorithms, you don't do encryption and signing of the message using the public-key algorithms themselves. Instead, you use traditional symmetric algorithms for encryption and hash ... View answer 0 votes If you really mean the public exponent, most likely, the exponent$e$is small; in fact, it's usually one of$\{ 3, 17, 65537 \}$. Just calculate$m^e \mod N$and check whether it equals$c$. If you'... View answer 0 votes I have seen this before in Java. Java's BigInteger class requires and generates binary data as signed little-endian. If the high bit of the first byte is set, the whole number is interpreted as ... View answer 0 votes For the problem of determining the base ($m$), the problem is that you don't have enough information. For any valid value of$e$, there is a matching value$m$that encrypts to the same$c\$; ...