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RSA: Theory and Implementation, Hacker News


RSA has been a staple of public key cryptography for over 40 years, and is still being used today for some tasks in the newest TLS 1.3 standard. This post describes the theory behind RSA – the math that makes it work, as well as some practical considerations; it also presents a complete implementation of RSA key generation, encryption and decryption in Go.

The RSA algorithm

The beauty of the RSA algorithm is its simplicity. You don’t need much more than some familiarity with elementary number theory to understand it, and the prerequisites can be grokked in a few hours.

In this presentationMis the message we want to encrypt, resulting in the ciphertextC. BothMandCare large integers. Refer to the Practical Considerations section for representing arbitrary data with such integers.

The RSA algorithm consists of three main phases: key generation, encryption and decryption.

Key generation

The first phase in using RSA is generating the public / private keys. This is accomplished in several steps.

Step 1: find two random, very large prime numbers (pandQand calculate . How large should these primes be? The current recommendation is fornto be at least 2048 bits, or over 600 decimal digits. We’ll assume that the messageM– represented as a number – is smaller than (n) (see Practical Considerations for details on what to do if it’s not).

Step 2: select a small odd integerethat is relatively prime to , which isEuler’s totient function. is calculated directly from Euler’s formula (its proof is on Wikipedia):

For wherepandqare primes, we get

In practice, it’s recommended to pickeas one of a set of known prime values, most notably65537. Picking this known number does not diminish the security of RSA, and has some advantages such as efficiency[1].

Step 3: computedas the multiplicative inverse of (e) modulo . Lemma 3 inthis postGuarantees thatdexists and is unique (and also explains what a modular multiplicative inverse is).

At this point we have all we need for the public / private keys. The public key is the pair and the private key is the pair. In practice, when doing decryption we have access tonalready (from the public key), sodis really the only unknown.

Encryption and decryption

Encryption and decryption are both accomplished with the samemodular Exponentiationformula, substituting different values ​​forxandy:

For encryption, the input isMand the exponent is (e) :

For decryption, the input is the ciphertextCand the exponent is (d) :

Why does it work?

Given (M) , we encrypt it by raising to the power ofemodulon. Apparently, this process is reversible by raising the result to the power ofdmodulon, gettingMback. Why does this work?


Recall thateanddare multiplicative inverses modulo. That is,. This means that for some integerkwe have  or.

Let’s see what is modulop. Substituting in the formula forEDwe get:

Now we can useFermat’s little theorem, which states that ifMis not divisible by (p) , we have. This theorem is a special case of Euler’s theorem, the proof of whichI wrote about here.

So we can substitute 1 for in the latest equation, and raising 1 to any power is still 1:

Note that Fermat’s little theorem requires thatMis not divisible byp. We can safely assume that, because if, then trivially  and again.

We can similarly show that:

So we have for the prime factors ofn. Using acorollary to the Chinese Remainder Theorem, they are then equivalent modulonitself:

Since we’ve defined (M) to be smaller thann, we’ve shown that  ∎

Why is it secure?

Without the private key in hand, attackers only have the result of , as well asnande(as they’re part of the public key). Could they inferMfrom these numbers?

There is noknowngeneral way of doing this without factoringn(see theoriginal RSA paper, section IX), and factoring is known to be a difficult problem. Specifically, here we assume thatMandeare sufficiently large that (otherwise decrypting would be trivial).

If factoring was easy, we could factornintopandq, then compute  and then finally finddfrom  using the extended Euclidean algorithm.

Practical considerations

The algorithm described so far is sometimes calledtextbook RSA(orSchoolbook RSA). That’s because it deals entirely in numbers, ignoring all kinds of practical matters. In fact, textbook RSA is susceptible toseveral clever attacksand has to be enhanced with random padding schemes for practical use.

A simple padding scheme called PKCS # 1 v1.5 has been used for many years and is defined inRFC 2313. These days more advanced schemes like (OAEPare recommended instead, but PKCS # 1 v1.5 is very easy to explain and therefore I’ll use it for didactic purposes.

Suppose we have some binary dataDto encrypt. The approach works for data of any size, but we will focus on just encrypting small pieces of data. In practice this is sufficient because RSA is commonly used to only encrypt a symmetric encryption key, which is much smaller than the RSA key size[2]. The scheme can work well enough for arbitrary sized messages though – we’ll just split it to multiple blocks with some pre-determined block size.

From (D) we create a block for encryption – the block has the same length as our RSA key:

PKCS #1 v1.5 encryption padding scheme

(Here) ************************ (PS) is the padding, which should occupy all the bytes not taken by the header andDin the block, and should be at least 8 bytes long (if it’s shorter, the data may be broken into two separate blocks). It’s a sequence of random non-zero bytes generated separately for each encryption. Once we have this full block of data, we convert it to a number treating the bytes as a big-endian encoding[3]. We end up with a large numberx, which we then perform the RSA encryption step on with. The result is then encoded in binary and sent over the wire.

Decryption is done in reverse. We turn the received byte stream into a number, perform, then strip off the padding (note that the padding has no 0 bytes and is terminated with a 0, so this is easy) and get our original message back.

The random padding here makes attacks on textbook RSA impractical, but the scheme as a whole may still be vulnerable tomore sophisticated attacksin some cases. Therefore, more modern schemes like OAEP should be used in

Implementing RSA in Go

I’ve implemented a simple variant of RSA encryption and decryption as described in this post, in Go. Go makes it particularly easy to implement cryptographic algorithms because of its great support for arbitrary-precision integers with the stdlibbigPackage. Not only does this package support basics of manipulating numbers, it also supports several primitives specifically for cryptography – for example the (Exp) method supports efficient modular exponentiation, and theModInversemethod supports finding modular multiplicative modular inverses. In addition, thecrypto / randcontains randomness primitives specifically designed for cryptographic uses.

Go has a production-grade crypto implementation in the standard library. RSA is inCrypto / RSA, so for anything realpleaseuse that[4]. The code shown and linked here is just for educational purposes.

The full code, with some tests, isavailable on GitHub. We’ll start by defining the types to hold public and private keys:

typePublicKeystruct{  N*big.(Int)    E*big.(Int)}typePrivateKeystruct{  N*big.(Int)    D*big.(Int)}

The code also contains aGenerateKeysfunction that will randomly generate these keys with an appropriate bit length. Given a public key, textbook encryption is simply:

FUNCEncrypt(Pub*PublicKey,(M)*big(Int))*(big).Int{  C:=(new)(big.(Int))  C.(Exp)((m),Pub.(E),Pub.(N))  returnC}

And decryption is:

FUNCdecrypt(priv*PrivateKey,C)*big.(Int))*bi g.Int{  M:=(new)(big.(Int))  M.(Exp)((c),priv.(D),priv(N))  returnM}

You’ll notice that the bodies of these two functions are pretty much the same, except for which exponent they use. Indeed, they are just typed wrappers around TheExpmethod.

Finally, here’s the full PKCS # 1 v1.5 encryption procedure, as described above:

// EncryptRSA encrypts the message m using public key pub and returns the// encrypted bytes. The length of m must be// otherwise an error is returned. The encryption block format is based on// PKCS # 1 v1.5 (RFC 2313).FUNCEncryptRSA(Pub*PublicKey,M[]Byte)([]Byte,error){  // Compute length of key in bytes, rounding up.  keyLen:=((pub).(N).(BitLen)   ()(7))/(8)    iflen((m))>keyLen-11{    returnnil,FMT.Errorf("len (m)=% v, too long",len(m)))  }  // Following RFC 2313, using block type 02 as recommended for encryption:  // EB=00 || 02 || PS || 00 || D  psLen:=keyLen-len((M))-(3)    eb:=make([]Byte,keyLen)  eb[0]=(0x)  eb[1]=(0x)    // Fill PS with random non-zero bytes.  fori:=(2);(i)(2)(psLen);{    _,err:=rand.(Read)((eb)[i:i1])    iferr!=(nil){      returnnil,err    }    ifeb[i]!=(0x){      i    }  }  eb[2psLen]=(0x)    // Copy the message m into the rest of the encryption block.  copy((EB)[3psLen:],(m))  // Now the encryption block is complete; we take it as a m-byte big.Int and  // RSA-encrypt it with the public key.  mnum:=(new)(big.(Int)).SetBytes((EB))  C:=encrypt(pub,(mnum))  // The result is a big.Int, which we want to convert to a byte slice of  // length keyLen. It's highly likely that the size of c in bytes is keyLen,  // but in rare cases we may need to pad it on the left with zeros (this only  // happens if the whole MSB of c is zeros, meaning that it's more than 256  // times smaller than the modulus).  padLen:=keyLen-len((C).Bytes())  fori:=(0);(i)(padLen)(i){    eb[i]=(0x)  }  copy((EB)[padLen:],C.Bytes())  return(EB),nil}

There’s alsoDecryptRSA, which unwraps this:

// DecryptRSA decrypts the message c using private key priv and returns the// decrypted bytes, based on block 02 from PKCS # 1 v1.5 (RCS 2313).// It expects the length in bytes of the private key modulo to be len (eb).// Important: this is a simple implementation not designed to be resilient to// timing attacks.FUNCDecryptRSA(priv*PrivateKey,(c)[]BYTE)([]Byte,error){  keyLen:=((priv).(N).(BitLen)()(7))/(8)    iflen((c))!=(keyLen){    returnnil,FMT.Errorf("len (c)=% v, want keyLen=% v",Len(c),keyLen)  }  // Convert c into a bit.Int and decrypt it using the private key.  CNUM:=(new)(big.(Int)).SetBytes((C))  mnum:=decrypt(priv,(CNUM))  // Write the bytes of mnum into m, left-padding if needed.  M:=make([]Byte,keyLen)  copy((m)[keyLen-len(mnum.Bytes()):],(mnum).(Bytes)())  // Expect proper block 02 beginning.  ifM[0](**********************************************************!=0x 00{    returnnil,FMT.Errorf("m [0]=% v, want 0x 00 "",(m)[0])  }  ifM[1]!=(0x){    returnnil,FMT.Errorf("m [1]=% v, want 0x 02 ",M[1])  }  // Skip over random padding until a 0x 00 byte is reached.  2 adjusts the index  // back to the full slice.  endPad:=bytes.IndexByte((M)[2:],(0x))(2)    ifendPad(2){    returnnil,FMT.Errorf("end of padding not found")  }  returnM[endPad1:],nil}

Digital signatures with RSA

RSA can be also used to performdigital signatures. Here’s how it works:

  1. Key generation and distribution remains the same. Alice has a public key and a private key. She publishes her public key online.
  2. When Alice wants to send Bob a message and have Bob be sure that only she could have sent it, she willencryptthe message with herprivatekey, that is. The signature is attached to the message.
  3. When Bob receives a message, he candecryptthe signature with Alice’s public key: and if he gets the original message back, the signature was correct.

The correctness proof would be exactly the same as for encryption. No one else could have signed the message, because proper signing would require having the private key of Alice, which only she possesses.

This is the textbook signature algorithm. One difference between the practical implementation of signing and encryption is in the padding protocol used. While OAEP is recommended for encryption, (PSS) is recommended for signing[5]. I’m not going to implement signing for this post, but the Go standard library has great code for this – for examplersa.SignPKCS1v 15andrsa.SignPSS.


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