# Tag Info

1

$d$ must indeed be an integer. To calculate $d$ you need to calculate $d=e^{-1}\bmod{\phi(n)}$ which is called the modular multiplicative inverse of $e\bmod{\phi(n)}$. For $d$ be computable you need to ensure that $$\gcd(e,\phi(n))=\gcd(e,(p-1)(q-1))=1$$ holds, which isn't the case with your sample parameters as $\gcd(3,60)=3\neq1$. As fgrieu pointed out ...

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The reason for both is that the generated values are trivial to detect / exploit and should be avoided and your RNG is deeply flawed if you actually get those values because the chance for this lies around $2^{-2000}$ if you use an appropriate parameter set. Now for the math: You need to choose your secret exponent $x$ such that $1\leq x\leq p-2$, with ...

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Because $a^{p-1} \equiv 1 \mod p$ and $a^{p-2} \equiv a^{-1} \mod p$. In both cases it is easy to solve for the exponent.

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I'll answer your question in order of appearance and leave the ones out which are off-topic here. For example I know you can get the $x/y$ or $x$ or $y$ from your public hexadecimal address, but can you get anything like that from the secret exponent? Not directly. You can use the secret exponent (a.k.a. private key) to calculate the public key ...

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This can be derived from two simple facts about the $mod$ operation: $a \bmod b = a + bi$ for some integer $i$ (for any $a, b$) $a \bmod c = b \bmod c$ if $a - b = ci$ for some integer $i$ With these two facts, we can look at $(g^a \bmod p)^b$; that can be simplified to $(g^a + pi)^b$ (for some integer $i$), and by the binomial expansion, this is $g^{ab} ... 0 It has to do with the Diffie–Hellman assumption. The DH key exchange is secure in groups where the computational DH assumption holds. One of the simplest such groups is the multiplicative group modulo a large prime. However, that is not necessarily required. At least some composite integers with unknown factors would make a secure Diffie–Hellman modulus, ... 2 Note: In my answer below I neglected to consider that the the messages for the associated signatures are also known, and that this could enable the existence of a practical algorithm to recover the modulus. While I haven't done the legwork to verify, fgrieu's comment below indicates that recovery of the modulus may very well be practical. I'm leaving my ... 3 calculate $$gcd(c_1^e - m_1 , c_2^e- m_2, \dots , c_k^e-m_k)$$ With a bit luck this should get$n\$. It will be an interesting exercise to calculate the probability of success based on the number of cleartect/ciphertext pairs.

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