Large data not encrypted with RSA Encryption

问题:

My Problem:

My encryption code working fine for below 64 characters. but if it exceeds 64 character I got following error

javax.crypto.IllegalBlockSizeException: input must be under 64 bytes

Encryption code

        cipher = Cipher.getInstance("RSA/ECB/PKCS1Padding);
        cipher.init(Cipher.ENCRYPT_MODE, publicKey);
        encryptedBytes = cipher.doFinal(message.getBytes(StandardCharsets.UTF_8));
        rsaEncrypted= Base64.encodeToString(encryptedBytes, Base64.NO_WRAP);

Key generation code

      KeyPairGenerator keyGen = KeyPairGenerator.getInstance("RSA");
      keyGen.initialize(512);
      KeyPair keypair = keyGen.genKeyPair();
      PublicKey pub = keypair.getPublic();
      byte[] pubs = pub.getEncoded();

My question:

It's possible encrypt large text with 512 bits of keys? Any mistake in my code?

Note: If anyone want full of code I will update later.

回答1:

Here is a direct quote from the seminal book titled Cryptography Engineering by Ferguson, Schneier, and Kohno,

Encrypting a message is the canonical application of RSA, yet it is almost never used in practice. The reason is simple: the size of the message that can be encrypted using RSA is limited by the size of n. In real systems, you cannot even use all the bits, because the encoding function has an overhead. This limited message size is too impractical for most applications, and because the RSA operation is quite expensive in computational terms, you don’t want to split a message into smaller blocks and encrypt each of them with a separate RSA operation.

In other words, for a n-bit RSA key, the maximum length of data RSA can encrypt in bytes is

Floor(n/8) - 11 

where 11 bytes is for padding

So for a key size of 512 bits, the maximum length of data that can be encrypted is,

512/8 - 11 = 53 bytes

Again from the book Cryptography Engineering,

The solution used almost everywhere is to choose a random secret key K, and encrypt K with the RSA keys. The actual message m is then encrypted with key K using a block cipher or stream cipher. So instead of sending something like E_RSA(m), you send E_RSA(K),E_K(m).

Basically, it's telling you do the following to get over the limitation of RSA,

1. Generate a secret key, K using an algorithm such as AES.
2. Encrypt the plaintext, m, with the newly generated secret key to get cipher text, say E_K(m).
3. Encrypt the secret key a RSA public key, E_RSA(K).
4. Sent the client the cipher text, E_K(m), and the encrypted key E_RSA(K).
5. The client can decrypt E_RSA(K) with the RSA private key to get K.
6. The client then decrypt the cipher text, E_K(m) with K to get m.

回答2:

The number of bytes you can encrypt in one RSA block is dictated by the key size used minus any bytes taken up for padding.

Generally RSA is not suited for bulk encryption as it's quite slow. Instead use a symmetric encryption algorithm like AES if you can. If you really need the two key's of RSA, use a Hybrid approach where you encrypt the data with a random symmetric key, and then encrypt that key with the RSA key.

A benefit of using symmetric encryption is also that the libraries automatically supports bulk encryption - which they don't for RSA.