Cracking RSA Ciphers That Have A Low-exponent

In RSA, we have:

The (mod N) part provides a core part of the security of the method. But when happens when M^e is less than N. Well, the cipher is easily cracked.

Normally, in RSA, we select two prime numbers of equal length (p and q), and then multiply these to give a modulus:

We then compute the cipher as:

But, if M^e≤N, then:

The message can be determined from:

For example, if e=3, we take the 3rd root of C. So, if we use a value of M^e that is less than N, and it is easy to recover the message with [here]:

m=libnum.nroot(C,e)

A sample run for a modulus of 128 bits is [here]:

Bob sends Alice a cipher using the RSA method. The cipher is 7446927644895231780144668992 and the public modulus is 114478088404552905363562514346363776111754070678655925650418939227582265249551089334182599044573960040630340706372893593053482627694097935698916862306537210729765728734386595471533687305757272884429400174867808817969676999360799494158594543289037055131414660873402975984872690402174487054555607828223347368227. The value of e is 3 and is a fairly low value. Using this information, can you crack the cipher?


Answer:
For e=3, we take the 3th root to get: 1952805748

Decipher…