Extra points 1

The last week I choosed explain the following expressions:

c = m^e mod n

and

m = c^d mod n

For this example I used  m = 123(message), n = 3233, d = 2753.

c = 123¹⁷mod 3233 = 855
m = 855²⁷⁵³mod 3233 = 123

The property that make RSA invertible is:

m^ed-1≡1 mod n

To get the message after encription we have de following relations(_ku -public key, _kr-private key)

D_kr(E_ku(m)) = (m^e mod n)^d mod n ⇔ m ≡ m^ed mod n
                                                          ⇔ 1 ≡ m^ed-1 mod n

123¹⁷ mod 3233 = 855 ⇔ 123 =123^17*2753 mod 3233 ⇔ 1 = 123^17*2753-1 mod 3233

We need to be sure that  the values of e, d, n satisfy:

∀ m ∈ Z : m^ed-1 ≡ 1 mod n

Now is possible note that the first two expressions work in noth directions.
When we have decryption first and then encryption, the expression is equal to:

E_ku(D_kr(c)) = c ^de mod n ≡ c mod n

855^2753*17 mod 3233 ≡ 855 mod 3233 = 855

This is the correctness in the two directions.


References

Applied Cryptography