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# RSA
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- Introduced in 1977 by Ron Rivest, Adi Shamir and Leonard Adleman
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- The most popular public key algorithm in the world
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- Solves an important problem that symmetric cryptography doesn’t
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- RSA keys are normally `2084` or `4096` bits
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- Security is built around the difficulty of *factoring large numbers*
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### RSA Encryption
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- Encryption performed by the *public key* can only be reversed using the *private key*
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### RSA Signatures
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- The authenticity of signatures generated by the *private key* can be verified by the *public key*
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### Euler Totient Function
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- Integers $a$ and $m$ are *relatively prime* if they do not share a divisor (except 1)
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- $gcd(a,m) = 1$
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- The **Euler totient** $\Phi$ is the number of integers in $\mathbb{Z}_m = \{0,1,...m-1\}$ for which $gcd(a,m)=1$
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- For example $\Phi(9)=6$ as:
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- $gcd(1,9)=1$ :white_check_mark:
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- $gcd(2,9)=1$ :white_check_mark:
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- $gcd(3,9)=3$ ❌
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- $gcd(4,9)=1$ :white_check_mark:
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- $gcd(5,9)=1$ :white_check_mark:
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- $gcd(6,9)=3$ ❌
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- $gcd(7,9)=1$ :white_check_mark:
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- $gcd(8,9)=1$ :white_check_mark:
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###
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#### Integer Factorisation
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- Any integer can be expressed as the multiplication of a list of prime numbers
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#### Calculating $\Phi(n)$
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- The totient is much easier to calculate given the prime factorisation of $n$
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$$
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m = p_1^{e_1}\cdot p_2^{e_2} ... \cdot p_3^{e_3} \\
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\Phi(n) = \prod^n_{i=1} (p_i^{e_i} - p_i^{e_i-1})
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$$
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##### $\Phi(p)$ for Primes
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$$
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\Phi(n) = \prod^n_{i=1} (p_i^{e_i} - p_i^{e_i-1}) \\
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\Phi(n) = (p^1 - p_0) = (p-1)
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$$
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This is similar for semi-primes $n=p\cdot q$
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$$
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\Phi(n) = (p^1 - p_0) \cdot (q^1-q_0) = (p-1)(q-1)
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$$
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#### Fermat’s Little Theorem
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- Fermat’s little theorem states that for some prime $p$, and any integer $a$:
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- $a^{p-1} \equiv 1 \space (mod \space p)$
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- Also note that $a^{p-1} = a\cdot a^{p-2} \equiv 1 \space (mod \space p)$
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- Therefore $a^{p-2}$ is actually the inverse of $a\space (mod \space p)$
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- It follows that $a^p \equiv p \space (mod \space p)$
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#### Euler’s Theorem
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- Generalisation of Fermat’s little theorem, not exclusive to primes
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- $a^{\Phi(m)} \equiv 1 \space (mod \space m)$
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- If $gcd(a,m)=1$
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- This works for any integer ring $\mathbb{Z}_m$
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- We can see that FLT is a special case of this
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- $\Phi(p) = (p-1) \therefore a^{\Phi(p)} = a^{p-1} \equiv 1 \space (mod \space p)$
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## RSA Key Generation
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1. Choose two large primes, $p$ and $q$
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2. Calculate the modulus $n=p\cdot q$
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3. Calculate $\Phi(n) = (p-1)\cdot (q-1)$
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4. Choose a value $e\in \{2, ..., \Phi(n) -1\}$ where $gcd(\Phi(n),e)=1$
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5. Compute $d$ where $d\cdot e \equiv 1 \space (mod \space \Phi(n))$
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$d$ is very easy to calculate if you know $p$ and $q$
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#### Example
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##### Encryption
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- Now we have a public key $(3, 187)$ and private key $107$
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- Encryption and decryption is performed by:
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- $x^e \equiv y \space (mod \space n)$
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- $y^d \equiv x \space (mod \space n)$
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#### Proof
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- We want to show that $(x^e)^d = x^{ed} \equiv x \space (mod \space n)$
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- Let’s assume $gcd(x,n)=1$ So Euler’s theorem applies
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- $e\cdot d=1\space (mod \space \Phi(n))$
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- $\therefore e\cdot d = 1 + k\cdot \Phi(n)$
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- $x^{e\cdot d} = x^{1+k\cdot \Phi(n)} = x\cdot x^{k+\Phi(n)}$
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- $x\cdot (x^{\Phi(n)})^k=x\cdot(1)^k=x$
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### Why is RSA Secure
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- We’d like the message $x$ based on some ciphertext $y$, given the public key $e$:
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- $y \equiv ?^d \space (mod \space n)$
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- $x \equiv y^? \space (mod \space n)$
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- It can be fairly easy to calculate $d$:
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- $e\cdot d \equiv q \space (mod \space \Phi(n))$
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- $\Phi(n) = (p-1)(q-1)$
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- As an attacker we only have access to $e$ and $d$
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### Exponentiation
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$$
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x^4 = x^2 \cdot x^2 \\
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x^8 = x^4 \cdot x^4
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$$
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When calculating a exponent raised to a power of two, we can use previously calculated values.
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##### Binary Exponentiation
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Where we treat the exponent as a binary number
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- We either square or multiply
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$26=11010_2$
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- Remember squaring is 1 bit shift to the left
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- Multiplying is just adding $1$
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$$
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x^{101} \quad = \quad x^{1100101_2} \\
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x\cdot x = x^2 \quad x^{10_2} \\
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x^2 \cdot x = x^3 \quad x^{110_2}\\
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x^3 \cdot x^3 = x^6 \quad x^{1100_2}\\
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x^6 \cdot x^6 = x^{12} \quad x^{11000_2}\\
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x^{12} \cdot x^{12} = x^{24} \quad x^{110000_2}\\
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x^{24} \cdot x = x^{25} \quad x^{110001_2}\\
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x^{25} \cdot x^{25} = x^{50} \quad x^{1100010_2}\\
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x^{50} \cdot x^{50} = x^{100} \quad x^{11000100_2}\\
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x^{100} \cdot x = x^{101} \quad x^{110001001_2}\\
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$$
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##### Computational Complexity
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- What is the computational complexity of exponentiation?
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- For a 2048 key:
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- $X^{2^{2048}}$ - A ridiculously big number
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- Where as using square and multiply
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- $2048=T$ we need $\frac{3T}{2}$ calculations
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