# Implementing RSA

This is some theory and guidance on implementing RSA cryptography and random prime generation. It was originally written for a school project about a year ago. In practice though, You probably shouldn't roll your own cryptography. If you'd like to see the full source code and writeup this was made with, it's here.

Asymmetric cryptography lets you encrypt a message using a key that is shared publicly, and then allows the encrypted message to be decrypted using a private key which can be kept secret. The private key can't be derived from the public key or from the message, meaning this is often used to accept messages from anyone who has your key without letting others read them.

The inverse also works, meaning you can encrypt a message with your private key and anyone with your public key can decrypt it. Whilst this isn't useful on its own, if you know what a message, or part of a message, should be, and you decrypt it to get that, you can know the sender does hold the private key. This is known as message signing.

## Theory

The following is not meant as complete proof of the security/functionality of RSA, rather as a general overview of why it is difficult to break and why it can be reliably reversed.

RSA relies on the fact that numbers are much easier to multiply than to factorise. This difference gets more pronounced the bigger the numbers are, meaning big enough numbers are almost impossible to factorise, but can still be multiplied relatively quickly. It can be shown that every non-prime (composite) number can eventually be factorised down to prime numbers, for example:

$2 * 2 * 5 * 5 * 5 = 2^2 * 5^3 = 500$

Each combination of primes is unique to one composite number, and every composite number, by definition, has a prime factorisation. This means if we take 2 primes and multiply them together, we know our original primes are the only prime factors of that number.

$n = pq$

This also makes it easy to calculate $$\phi(n)$$. Since p and q are prime, from Euler's Totient Theory 2:

$\phi(pq)=(p-1)(q-1)$

We now choose $$e$$ (not the constant), which is coprime to $$\phi(n)$$. Then we find $$d$$ such that $$de \equiv 1(\mod \phi(n))$$, which we can do using the extended Euclidean algorithm.

This gives us the public key: $$(e, n)$$ and the private key: $$(d,n)$$.

If we have a number $$M$$ to encrypt ($$M< n$$), we simply use these functions:

\begin{align*} E(M) &= M^e (\mod n) \\ D(C)&=C^d (\mod n) \end{align*}

For our uses, signing a number $$M$$ looks like this:

$S(M) = M^d (\mod n)$

And if the following is true, $$C$$ was encrypted using the private key:

$C^e (\mod n) = M$

## Finding p and q

Proving a large number to be prime is far too costly to generate keys, so instead tests that give a high probability of a given number being prime will be used, and random numbers generated till we get 2 that have a high enough probability.

Since every prime takes the form $$6n + 1$$ or $$6n -1$$ (except 2 and 3), we can generate $$n$$ randomly and put it in the above equations until we generate our 2 primes.

Fermat's Theorem dictates that if $$x$$ is prime, $$a^x - 1 \equiv 1 (\mod x)$$ for any values of $$a$$ less than $$x$$3. However, it's possible for a number to pass this test even if it isn't prime, so another test is needed.

It can be shown that $$x$$ is prime if and only if $$n^2 = 1 (\mod x)$$ only when \$n = ± 14. So if we can find $$n$$ such that $$n^2 (\mod x) \neq \pm 1,n < x$$ , we know that $$n$$ is not prime. If we take $$n^{2}=a^x - 1$$, where $$a$$ is from our previous Fermat Test, we know that if $$x$$ is prime it's also odd, and thus $$x-$$ is even. This means we can also write $$a^x - 1 = a^{{2}sd}$$ (d is odd). This means that for the sequence:

$a^{{2}sd} \mod x, a^{{2}{s-1}d} \mod x, a^{{2}{s-2}d} \mod x, ..., a^{d} \mod x$

Each number is a square root of the previous number. If $$x$$ is prime then for this sequence, either every number is 1, or the first number that isn't 1 is $$x - 1$$.

These 2 techniques can be combined to generate numbers that are likely to be prime at a relatively low computational cost. Since this will only be used for key generation, it's not as important that this function is fast.

## Finding e

e being (relatively) small makes encryption more efficient without affecting security.

Most commonly, e is taken as 65,537 and if this isn't coprime to $$\phi(n)$$, we choose a different p and q. This is because it's a relatively large known prime which can fit well within 32 bits.

An alternative way to do this would be to loop from $$t$$ up to $$\phi(n)$$ and find the first integer where $$gcd(e, \phi(n)) = 1$$.

## Finding d

d can be calculated using the extended Euclidean algorithm5, which we can use to solve $$ae + bn = gcd(e, n)$$. Since the two are coprime, $$gcd(x,y) = 1$$. If we apply $$(\mod n)$$ to the whole equation, the $$bn$$ disappears so we're left with $$ae = 1(\mod n)$$. This means $$a = d$$.

For RSA to work, Its required that the message $$M < n$$. An easy way to ensure this would be to make sure the number of bits for $$M$$ is less than $$n$$. If the message is longer than that length, we can split it into many smaller parts and encrypt these parts individually. For convenience, A padding function will be used to make sure every message is the same length of bits.

OAEP is the standard padding scheme specified by PKCS#1 and used by most implementations. This requires a function for generating random bytes as well as a hashing function. If the final $$M$$ passed to the RSA signing function is $$k$$ bits long, half of $$k$$ will be allocated to the actual message and the other half to the random seed. This means $$k$$ is twice the length of the output from the hash function used.

If a message spans over more than one padded 'block', then the same seed number should be used by each block. This will prevent being able to 'swap' blocks from other encrypted messages into different ones, as the seed number will likely be different from the rest of the 'blocks'.

## Size of numbers

The generally accepted size for the modulus $$n$$ is 2048 bits. Since we're using a constant for $$e$$ we know that its only 32 bits. $$d$$ is likely to be closer to the size of the modulus, so we'll assume its max is the same as $$n$$.

Max $$n$$ and $$d$$: $2^{2048} − 1 \approx 3.23 * 10^{616}$

Max e:

$2^32 − 1 = 4,294,967,296$

(They are all unsigned integers)