I am a programmer, While I was studying the RSA encryption I found some difficulties with some mathematical matters

RSA algorithm has the following concepts

Modulo, Modular multiplicative inverse, Prime numbers, Totient $ϕ$

For me I understand all these concepts but to make my question clear let me take an example

The steps to encrypt some text are :

First find two prime numbers

Lets take $5$ and $7$

Then find the modulus by multiplying these numbers $5*7 = 35$

Now find the totient $ϕ \implies ϕ(5*7) = (5-1) * (7-1) = 24$

Now to fin the public key I have to find a number which has GCD with the totient equals to one which is $11 \implies \gcd(11,24) = 1$

The private key is the modular multiplicative inverse of $11$, in this case $107$ is the multiplicative inverse of $11$ because $11*107 \equiv 1 \pmod{24}$.

Now we have the public key $11$ and the private key is $107$.

Lets denote the the public key by $e$ and the private key by $d$ Now If we want to encrypt a message the equation is $c \equiv m^e \pmod n$

and if we want to decrypt the message the equation is $m \equiv c^d \pmod n$

Quick test let's decrypt $m = 5$

$5^{11} \bmod 35 = 10 \implies c= 10$

$10^{107} \bmod 35 = 5 \implies m= 5$

this is magic, actually I understand the concepts but I do not understand how is that done, I understand that $d$ is the inverse of $e$ because we got it using the modular multiplicative inverse but I do not understand why the modulus is $n$ and not $ϕ$ I do not understand why the public key must have $gcd$ equals $1$ with $ϕ$ not with $n$, I am confuses and I need your help to explain that to me.


I think the "magic" you are looking for is

Eulers Theorem

which states that $a^{\phi(n)} \equiv 1 \mod n$ if $\gcd(a,n)=1$.

  • $\begingroup$ I am a programmer not a mathematics student so I do not know a lot about these things, thanks a lot for the reference. $\endgroup$ – Mohamad Sep 12 '15 at 23:48
  • $\begingroup$ Euler's theorem is not enough here, because RSA also works when the message is not coprime to the modulus (because it might be a multiple of one of the secret primes). This is a particular case for square-free moduli, which you can prove using only Fermat's little theorem and the Chinese Remainder Theorem. $\endgroup$ – Henning Makholm Sep 12 '15 at 23:53

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