# In the RSA algorithm, what is the security flaw with choosing e = 1 for the public key?

Quote from Raj Maurya on February 23, 2019, 7:52 amIn public key cryptosystems, the RSA is the most commonly used algorithm. Answer the following questions: -

1. In the RSA algorithm, what is the security flaw with choosing e = 1 for the public key? (Hint: the RSA formula for encryption is c = M e mod n).

2.You are given a cipher text C=10 by a user whose public key is e=5, n=35. Break the code.

Solution1)

When e = 1

c = M ^ e mod N

c = M mod N

We know that N can be known by public

so message easily decrypted

2)

when c = 10, n=35, e=5

c = M^e mod 35

10 = M^5 % 35

when

M = 1 => 1^5 %35 = 1 not equal to 10

M = 2 => 2^5 %35 = 32 % 35 = 32 not equal to 10

M = 3 => 3^5 %35 = 81 % 35 = 21 not equal to 10

M = 4 => 2^5 %35 = 256 % 35 = 11 not equal to 10

M = 5 => 5^5 %35 = 3125 % 35 = 10 equal to 10

So M = 5

In public key cryptosystems, the RSA is the most commonly used algorithm. Answer the following questions: -

1. In the RSA algorithm, what is the security flaw with choosing e = 1 for the public key? (Hint: the RSA formula for encryption is c = M e mod n).

2.You are given a cipher text C=10 by a user whose public key is e=5, n=35. Break the code.

**Solution**

1)

When e = 1

c = M ^ e mod N

c = M mod N

We know that N can be known by public

so message easily decrypted

2)

when c = 10, n=35, e=5

c = M^e mod 35

10 = M^5 % 35

when

M = 1 => 1^5 %35 = 1 not equal to 10

M = 2 => 2^5 %35 = 32 % 35 = 32 not equal to 10

M = 3 => 3^5 %35 = 81 % 35 = 21 not equal to 10

M = 4 => 2^5 %35 = 256 % 35 = 11 not equal to 10

M = 5 => 5^5 %35 = 3125 % 35 = 10 equal to 10

So M = 5