# Thread: show encryption and decryption are identical

1. ## show encryption and decryption are identical

show that the encryption procedure C=P^e(mod n) and decryption procedure P=C^d(mod n) in RSA cryptosystem are identical when encryption is done using modular eponentiation with modulus n=35 and enciphering key e=5

2. They will be identical as your encryption and decryption key are identical for the following reasons;

If e=5 and mod m= 35, then in order to get your decryption key you need to find

$\displaystyle \frac{1}{5} mod\ \phi(35)$

Since this key will be the same as the encryption key, encrypting and decrypting will use the same exponent, not very secure.

3. Originally Posted by mandy123
show that the encryption procedure C=P^e(mod n) and decryption procedure P=C^d(mod n) in RSA cryptosystem are identical when encryption is done using modular eponentiation with modulus n=35 and enciphering key e=5
Just remember that d and e should satisfy $\displaystyle de = 1\text{ mod }\phi(35)$
And $\displaystyle \phi(35) = (7-1)(5-1) = 24$. Now you will have to solve for
$\displaystyle 5d = 1\text{ mod }24$. Clearly $\displaystyle d=5$

Thus $\displaystyle d=e=5$ and hence encryption and decryption keys are identical

4. Hello,

Originally Posted by jtsab

$\displaystyle \frac{1}{5} mod\ \phi(35)$
Is this writing correct ? Oo

5. Originally Posted by Moo
Hello,
Is this writing correct ? Oo
There is no harm, if he means $\displaystyle 5^{-1} \mod \phi(35)$, right?

6. Yep, but the writing still looks weird for me

7. Originally Posted by Isomorphism
There is no harm, if he means $\displaystyle 5^{-1} \mod \phi(35)$, right?
That's what I was implying, since $\displaystyle \frac {1}{5} =\ 5^{-1} \mod \phi (35)$