LAFlMPvyvmvaymaUvQvEv9v-v!v9tel?coprimesLambda%lambda-
".:FR^jv?End-to-End Encryption"These are the numbers so far:
P =
Q =
N =
λ(N) =
E =
D =
All numbers are kept secret, except for E and N, which are needed by Bob to encrypt the message (password / key).
Alice sends N and E to Bob.
()
Press continuprimePprimeQmodulus-varEvarDposs
alicesmessageContinu Continu O?End-to-End Encryption"These are the numbers so far:
P =
Q =
N =
λ(N) =
E = r
We need to find 1 more number: D (for Decrypt).
There's one rule:
(E × D) mod N must be 1
(E × D) mod N = 1
( × ?) mod = 1
Again, this is done through trial and error. Pick a number, and check if it's a match. If not, pick another number and try again.
We settle on the number:
D = , because:
( × ) mod F = 1
Again, there were multiple options to choose from.
Press continuWY[-]][_]_[Continu Continu M?End-to-End Encryption"These are the numbers so far:
P =
Q =
N =
λ(N) =
We need to find 2 more numbers, E (for Encrypt) and D (for Decrypt).
For E you'll have to pick a number, and check if it's a match. If not, pick another number and try again.
Three rules for this:
• E and λ(N) are coprime.
• E must be prime.
• E must be between 1 and λ(N) ( )
We settle on the number:
E = ;
There were multiple options to choose from.
Press continuWY[--]Continu Continu K?End-to-End Encryption]Alice picks two random prime numbers, that are different numbers but of similar magnitude:
P
Q 7
She uses these numbers to calculate N:
N= P × Q
N = ×
N = x
Now she needs to calculate the Carmichael's totient function of the product as:
λ(N) = lcm(P − 1, Q − 1)
λ(
) = lcm((
− 1) × (
− 1))
λ( ) = lcm( × )
λ() =
Press continuWYWY[[WY[W?Y?[-ContinuContinu alicesfilesWY[-]_a[]a:9_?__]
remainder1?verheffing1--]_hekkie?_varEcoprimes]33coprimevalueprimevalue2primes%@coprimeindex-@W?Y?-WY[WYY@Y@&@YW@W@&@W3vvvttvvv@@?notprime
primevalue@@@@vvvvvvvAlicenamew?End-to-End Encryption Bob receives:
(N, E)
([)
He uses these numbers in a formula to encrypt his private key M, turning it into C:
M =
C = (M^E) mod N
C = (^) mod
C = () mod
C = )
Bob sends C to Alice, keeping M secret.a
privatekeya?averheffing2avarCContinuContinu bobsfilesbobsmessageaaa?a12BobvarM?End-to-End Encryption!Alice receives:
(Encrypted) C = (P)
She uses this number, and her own numbers to decrypt C.
M = (C^D) mod N
M = (^) mod
M = mod
M =
Press continu @@verheffing3@ContinuContinu@@@Alice?End-to-End EncryptionCAs you can see, the numbers are identical!
Bob's unencrypted key:
Alice's decrypted key:
The encryption is one way.
You can only decrypt the key with the decryption number, calculated earlier.
If you want to use letters, you can set a number to each letter, and convert it that way.
Because this is a pretty slow process, messages aren't often encrypted with RSA, just the key for the regular (end to end) encryption.DoneDone?End-to-End Encryption|Somewhere along the line, a calculation provided a number too big for Automate, so the numbers don't match.
Alice's key:
><
Bob's key:
>
<
Try again?ManualManual dad3child3dad2child2dad1child1?End-to-End EncryptionAlice wants to send a message to Bob, using end to end encryption.
In order to do that, they must establish some sort of mutual encryption key. A widely used way to do this is a key exchange, using a RSA encrypted key.ContinuContinu##############################
57=;9I? KMOGCAE