Let $d$ be the inverse of $e \bmod \phi(N)$. Therefore, $de \equiv 1 \pmod{\phi(N)} $ $d \equiv e^{-1} \pmod{\phi(N)} $ In RSA encryption, we can decrypt messages because $P^{ed} \bmod N = P^{1} \bmod N$ I don't understand why $P^{ed} \bmod N = P^{1} \bmod N$ if $e$ and $d$ are modular inverses of $\phi(N)$. What property is this?
81102
As you know, in RSA encryption, we calculate an n value by multiplying two long primes. Then we calculate Φ(n), which is equal to (p-1)(q-1). After that, we choose an encryption key which satisfies gcd(e,Φ(n))=1. I know that it is necessary to calculate decryption key, which is the modular inverse of e. My question is: Why do we use Φ(n) in determination of encryption and decryption key processes? How can one prove it is working for RSA encryption?
47760
I am looking for ways to prove that $\mathbb{Z}_m=\{0,1,2,\dots,m-1\}$ is a field under addition and multiplication $\bmod\ m$ iff $m$ is a prime.   I tried it this way: If $m$ is a prime $p$,it is obvious that most importantly, we need to prove that every non-zero element in $\mathbb{Z_p}$ has an inverse in $\mathbb{Z_p}$.We note that if $a,b\in \mathbb{Z_p}$, then $a\cdot b\in \mathbb{Z_p} \dots \boxed{*}$. Next, $a^p\equiv a \bmod p$ for prime $p$(the proof follows by induction on $a$).As  $\displaystyle 0\leq a\leq p-1$,$a\not =p$. Thus $$a^{p-1}\equiv 1 \bmod p$$ for prime $p$ which,together with $\boxed{*}$ guarantees the existence of an inverse in $\mathbb{Z_p}$.Proving the other properties is simple.  Conversely,if $\mathbb{Z_m}$ is a field,then every non-zero element of $\mathbb{Z_m}$ has an inverse.Note that inverse of an element is unique.For if $a\cdot b\equiv a\cdot c\equiv 1 \bmod m$,then $(a\cdot b)\cdot c\equiv c\equiv (a\cdot c)\cdot b\equiv b \bmod m$ i.e $m|(b-c)\implies b-c=0(0\leq a\leq m-1,a\in \mathbb{Z_m})$.So, inverses must be unique and every element is either an inverse of itself or has in inverse in the field.That allows us to conclude that $[(m-1)!]^2\equiv 1 \bmod m$ which can only mean that either  $(m-1)!\equiv 1 \bmod m$ or $(m-1)!\equiv -1 \bmod m$. If $m$ is not a prime, it has a prime factor, say $p$. $ 0\leq p\leq m-1\implies p|(m-1)!\implies p|1$or $-1$ which is absurd, a contradiction.Thus $m$ is prime.  I feel that this method is too involved and complicated. Can anyone please suggest some other method?(Some context: This is a problem from Finite Dimensional Vector Spaces by Paul Halmos. I am learning linear algebra for the first time.It seems some of the comments have been posted on the assumption that I know rings and fields; I don't. )
24393
I have the following equation: $$ n \equiv M^k\mod(b) $$ where n, a, b are integers, and M is unknown. How do I solve this equation to find the M value? Those a and b are public keys of the RSA algorithm and n is the result of encryption. I need to find the encrypted message.
117732
Question 1: Denote $a \mod b$ as $a \% b$, where $a$ and $b$ are some integers  Evaluate 12^32475 % 21    The following is what I tried:    12^32475 % 21 = (12^3 % 21)^10825 % 21 = 6^10825 % 21 = ... = 6^433 % 21   But how to continue ?  Question 2:  I don't understand this: n-1= -1 mod n. Where is this from? And does this line become the next line?    Context: my class is discrete math for computer science. So far for related topics, it covers modular inverse, (extended) GCD algorithm, Fermat's Little Theorem and Chinese Remainder Theorem (CRT).
79879
I would like to factorise $496241$. I know the answer is $677 \times 733$. But I don't know how to get there.   Here is the full question:   "A message has been encoded using RSA with a modulus of $m = p q = 496241$ (with $p$ and $q$ being primes) and encoding exponent of $218821$.  You are advised that $\{13631, 142703\}$ is a valid encoding–decoding pair for the same modulus, $m$.  (a) Use this information to determine $\phi(m)$ for this modulus. (Using software to directly factorise $m$ is not a valid option for doing this part.)  (b) Verify your answer by determining the primes $p$ and $q$. Show how these combine to give both $m$ and $\phi(m)$."  I need it so I can solve (a), and it says I cannot use software to directly factorise m. So I guess pen and paper. But there are not tutorial online. Thanks
44584
I decided to read the original RSA paper A Method for Obtaining Digital Signatures and Public-Key Cryptosystem because of a question I had about RSA (which is not the question I'm about to ask, but may be a question on here at some time) and came across something I had never seen before in any textbook description of RSA (Note: this comes from the beginning of ).  for any integer (message) $M$ which is relatively prime to $n$, $M^{\phi(n)} \equiv 1\pmod{n}$  The part I don't recall ever hearing in class, or coming across in a text book, or seeing in any description of RSA I've ever read is that $M$ and $n$ must be relatively prime. So, I decided to play around with a toy example of RSA to see what would happen. I use $p=13$, $q=31$, $e=7$ and $M=2p=26$ (for completeness: $d=103$, $\phi(n)=360$, $n=403$). I observed the following using my toy instance of RSA: $M^e\equiv M\pmod{n}$, $M^e\equiv 0\pmod{p}$ and $M^e\equiv M\pmod{q}$. Do these equivalences hold for all $M,p,q$ when $M$ and $n$ are not relatively prime? How would I prove this? What is the real-world effect of this?
111629
While studying the RSA algorithm I referred to some books and some sites such as  and all of them chose {d,n} as the secret (private) key and release {e,n} as the public key but as d and e are multiplicative inverse of each other, can't we keep {e,n} as private key and use the other as public key? Is it a standard rule to choose the {d,n} as the private key?
36597
I would like to prove the following equality: $$k^{\phi(l)} + l^{\phi(k)} \equiv1\pmod{lk}$$if $\gcd(l,k)=1$. What methods can I use? Thank you for your help.
64288
How does $a\equiv b\pmod n$ implies $a$ and $b$ have the same remainder when divided by $n$?     I don't understand the huge jump from modular equivalence to having equal remainders.   I see by definition  $a\equiv b\pmod n\,$ implies $n\mid (a-b)\,$ so $\,a-b=nk\,$ so $\,a=b+nk$.     But I do not see how this implies that $a$ and $b$ have the same remainder when divided by $n$.
63500
It's using Euclidian inverse. Can you show it step by step?   $$N=7, q=11, a = 3+2X^2 -3X^4 +X^6$$  the inverse of $a \pmod {11}$ is  $$A=-2+4X+2X^2 +4X^3 -4X^4 +2X^5 -2X^6$$  How do we compute $A$?
118796
For example, in RSA, we use this for encryption: $ciphertext = (m^e \mod n)$  and for decryption.  If our message is "hello world", then what number do we have to put as $m$ in the RSA formula?
27820
In an Introduction to Abstract Algebra by Thomas Whitelaw, he gives examples of the congruence mod operation, such as $13 \equiv5 \pmod4$,  and $9 \equiv -1 \pmod 5$. But when I first learned about the modulo operation my junior year, I would have told you that $13 \equiv 1 \pmod 4$, and that $9 \equiv 4 \pmod 5$.     Is this just a difference in the definition of modulo? Or is this pretty typical (to not take it to the largest factor, or one over)?
30477
I was curious what's to happen when $p = q$ for $N=pq$ in RSA scheme.  First I realize that one can easily find out $p$ and $q$ by taking a square root of $n$.  However, it appears to me that under $\pmod{N}$ where $p = q$, one cannot reliably retrieve the original message via $m^d$ where $m$ denotes an encrypted message and $d$ denotes the decryption key. Can someone explain why?  Thank you.
14976
I'm going to encrypt the characters Zhu, and decrypt them using RSA. I'm using the public key $\{e, n\}$ and private key $\{d, n\}$. The values of $e$, $d$ and $p$ I get from my textbook: $e = 17$, $d = 53$, $n = 77$.  It works fine when the number is smaller than 76, but in my example below, it doesn't work:  z = 90  h = 104  u = 107  e = 17 d = 53 n = 77  print z,h,u  c_z = z ** e % n c_h = h ** e % n c_u = u ** e % n  print c_z, c_h, c_u  p_z = c_z ** d % n p_h = c_h ** d % n p_u = c_u ** d % n  print p_z, p_h, p_u   The outputs are:  90 104 107 62 69 46 13 27 30     Do you have any idea what's going wrong?
92411
If I have an encoding function  $f(x)$ that maps a message $m$ to a point $P$ on a suitable Elliptic Curve $E$ . If I have the public key $Q$ of my recepient then I can encrypt the message as follows:  Choosing a random number $k &lt; n -1 $ where $n$ is the order of the curve E .  Calculating $C = [k] G $ where $G$ is the generator point of the curve E   Calculating $R = [k]Q $ where $Q$ is the recipient's public key   Now the message is encrypted by adding $P$ to $R$ to get          $C_e$ $( = R + P)$ which is sent along with $C$ to the recepient.  DECRYPTION:  Now for the recepient to decrypt my message they compute $R = [privatekey] C $ (where $C = [k]G $) and simply subtract $R$ from $C_e$ to get the encoded point $P$ back which is then decoded by an inverse function $f^{-1}(x)$ to recover $m$.  Now , two problems arise : If the attacker knows the plaintext (suppose if I follow a particular format of data while sending messages ) he could encode his guessed plaintext and subtract it from $C_e$ to recover $ R $ back ! . If I used that same $ R  $ to encrypt further "blocks" of my message then the security of the later parts of my message has been breached!   I realize that doing  something like a scalar multiplication over and over again for every block of data, would be a huge drawback in efficiency and speed.   Is there any way to compute a different $ R $ for every block of data, without using much resources and that too quickly ?   Does reusing the same $R$ for another message break security ?  NOTE: This question is for educational purposes only. It's for the sake of expanding my knowledge on Crypto.
38265
I need to find out the modular inverse of 5(mod 11), I know the answer is 9 and got the following so far and don't understand how to than get the answer. I know how to get the answer for a larger one such as 27(mod 392) but am stuck because they are both low numbers.  11=5 (2)+1  5=1 (5)
63331
So I know that if $(a,n)=1$, then $a^{\phi(n)}≡1\pmod{n}$ by Euler's Theorem. I also know that $d \mid a^k$ and obviously $d \mid n$. I know I probably need to use this in the proof, but I just don't know how to get started.