How to solve RSA Algorithm Problems?

RSA algorithm is an asymmetric cryptography algorithm which means, there should be two keys involve while communicating, i.e., public key and private key. There are simple steps to solve problems on the RSA Algorithm.

Example-1:

Step-1: Choose two prime number $p$ and $q$
Lets take $p = 3$ and $q = 11$
Step-2: Compute the value of and
It is given as,
```
 and 
```
Here in the example,
$n = 3 \times 11 = 33$
$\phi = (3-1) \times (11-1) = 2 \times 10 = 20$
Step-3: Find the value of $e$ (public key)
Choose $e$ , such that $e$ should be co-prime. Co-prime means it should not multiply by factors of $\phi$ and also not divide by $\phi$

Factors of $\phi$ are, $20 = 5 \times 4 = 5 \times 2 \times 2$ so $e$ should not multiply by $5$ and $2$ and should not divide by 20.

So, primes are 3, 7, 11, 17, 19…, as 3 and 11 are taken choose $e$ as 7

Therefore, $e = 7$
Step-4: Compute the value of (private key)
The condition is given as,
where y is the value of .

To compute the value of $d$ ,
1. Form a table with four columns i.e., a, b, d, and k.
2. Initialize a = 1, b = 0, d = $\phi$ , k = – in first row.
3. Initialize a = 0, b = 1, d = $e$ , $k = \frac{\phi}{e}$ in second row.
4. From the next row, apply following formulas to find the value of next a, b, d, and k, which is given as
  - $a_{i} = a_{i-2} - (a_{i-1} \times k_{i-1})$
  - $b_{i} = b_{i-2} - (b_{i-1} \times k_{i-1})$
  - $d_{i} = d_{i-2} - (d_{i-1} \times k_{i-1})$
  - $k_{i} = \frac{d_{i-1}}{d_{i}}$
As soon as, $d = 1$ , stop the process and check for the below condition
```
if 
    
if 
    
```
For a given example, the table will be,

a b d k

1 0 20 –

0 1 7 2

1 -2 6 1

-1 3 1 –

As in the above table $d = 1$ , stop the process and check for the condition given for the $b$
$\therefore b = 3$

To verify that $b$ is correct, the above condition should satisfy, i.e.
$gcd(\phi, e) = \phi x + ey = (20 \times -1) + (7 \times 3) = 1$ . Hence $d$ is correct.
Step-5: Do the encryption and decryption
Encryption is given as,
$c = t^{e}\mod n$
Decryption is given as,
$t = c^{d}\mod n$

For the given example, suppose $t = 2$ , so
Encryption is $c = 2^{7}\mod 33 = 29$

Decryption is $t = 29^{3}\mod 33 = 2$

Therefore in the final, $p = 3$ , $q = 11$ , $\phi = 20$ , $n = 33$ , $e = 7$ and $d = 3$

a	b	d	k
1	0	20	–
0	1	7	2
1	-2	6	1
-1	3	1	–

Example-2: GATE CS-2017 (Set 1)
In an RSA cryptosystem, a particular A uses two prime numbers p = 13 and q =17 to generate her public and private keys. If the public key of A is 35. Then the private key of A is?

$p = 13$ and $q = 17$
Compute $n = 13 \times 17 = 221$ and $\phi = (13-1) \times (17-1) = 12 \times 16 = 192$
$e = 35$ (public key)
Compute $d$ (private key)

a b d k

1 0 192 –

0 1 35 5

1 -5 17 2

-2 11 1 –

$\therefore d = 11$ (private key)

a	b	d	k
1	0	192	–
0	1	35	5
1	-5	17	2
-2	11	1	–

My Personal Notes

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