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The attack is even more simple with RSA than with symmetric keys, because the asymmetric encryption key is assumed to be public. Let me tell you a story involving Alice, Bob and Mallory :). Alice wants to send a message to Bob using RSA. Alice encrypts the message using Bob's public key and sends it Mallory performs a Man-In-The-Middle attack, and ...


4

How did they jump from the spectrogram showing the RSA exponentiation timings straight to the secret bits? Actually, they're jumping to secret bits; however those aren't the secret bits you're thinking of. The bits displayed above are not the actual bits of $p$ and $q$. Instead, those are the bits from the secret exponent; because GnuPG uses CRT (and ...


2

Two scenarios: Alice sends a document to Bob that she has signed using her private key. Bob verifies the signature. Alice sends a document to Bob. Bob encrypts a random message for Alice's public key. Alice decrypts it and sends it back to Bob. Bob verifies it's the same. Both will only be possible if Alice has access to the private key, so you might ...


2

I'm assuming you mean a base 64 encoded key file, since removing the newlines from a binary file would obviously break things. The RSA standards (e.g. RFC 2459) only define a binary representation for keys. In practice, like OpenPGP keys (RFC 4880), they are often encoded in base 64 using the otherwise obsolete PEM standards (RFC 1421). The PEM printable ...


2

In practice, no. Firstly, RSA private keys are typically stored together with the public exponent. The standard definition includes the following information: RSAPrivateKey ::= SEQUENCE { version Version, modulus INTEGER, -- n publicExponent INTEGER, -- e privateExponent INTEGER, -- d prime1 INTEGER, -- p ...


2

Your javascript library linked to has no restrictions on key size. Many libraries out there that implement RSA will have a restriction on the key size. This is to make sure developers are following best practices as if the key size is too small, the security of the cipher is completely blown. It looks like the Java library you are using won't let you use key ...


1

Theoretically there is no requirement for the public key to be hard to guess given the private key. The public key is assumed to be known by all parties, including the adversary, so there is usually not much point in making it hard to guess. In fact, the question depends a lot on what you define as the private and public keys: For example it is not ...


1

We all know that in a public key cryptosystem, given a public key it is extremely hard to compute private key from it. Is it the same case in reverse? Not in general. For example, in Diffie-Hellman the public key is just a constant raised to a private number, modulo another constant. In some elliptic curve algorithms, the public key is a curve ...


1

No. An RSA signature is just a single number, encoded in a certain way. The number represents $x^d$, where $x$ is a padded hash of the document and $d$ the private exponent. If you know a public key $(m,e)$, you can calculate $x = (x^d)^e \mod m$, but without a document (or at least its hash) there is nothing to compare it with to verify anything. The only ...


1

References Related to your reference request: SHA512withRSA points to the RSA Signature Scheme with Appendix based on PKCS #1 v1.5 with SHA-512 hash function. This means you’re looking for reference documentation describing RSA PKCS1 v1.5 (see: RFC2313) signatures with SHA512 (see: RFC6234) hash and X.509 encoding format. Removing “overhead” from code As ...


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You've got the vague idea about it, but you've mixed up some terms (you say Bob uses Amy's private key in your first paragraph when I think you meant to say public key; if Bob has Amy's private key it's all over for her). Also, if it was Bob sending Amy the message, Amy would be trying to verify it was truly Bob who sent her that message, not the other way ...



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