# Affine cypher. Find function and plaintext

I was checking the following Affine Cipher / modular aritmethic exercise:

You intercept a ciphertext YFWD, which was ciphered using an affine cipher. You know that the plaintext starts in ST, find the cipher function and the plaintext $$\pmod{26}$$

I know $$Y → S$$ and $$F→T$$ , also $$Y=24,S=18,T=19,F=5$$

I've been trying to start from a congruence's equation system like this one:

$$\begin{cases}25 & \equiv & 18a+b\pmod{26}\\ 5 &\equiv & 20a+b \pmod{26}\end{cases}$$

From this point I can't find a way to solve the system, so any help will be really appreciated.

• How would you solve the system if it were $25=18a+b$, $5=20a+b$? Do exactly that, but doing all the arithmetic modulo 26. – Gerry Myerson Feb 6 at 4:58
• And note that 20 and 18 have no inverses in $\pmod{26}$. You will have multiple solutions to eliminate. – kelalaka Feb 6 at 8:41

Subtracting $$\begin{cases}25 \equiv & 18a+b\pmod{26}\\ 5 \equiv & 20a+b \pmod{26}\end{cases}$$ gives $$-20\equiv 2a\pmod{26}$$, from which $$a\equiv -10\equiv 3\pmod{13}$$, that's $$a\equiv 3\pmod{26}\lor a\equiv 16\pmod{26}$$. In both cases, we get $$b\equiv 23\pmod{26}$$.

Consequently, the pairs $$(a,b)$$ of solutions modulo $$26$$ are: \begin{align} &(3,23)&&(16,23) \end{align}

Your equations are not quite correct: if the encryption function is $$E(x) = ax+b \pmod{26}$$ where $$x \in \{0,\ldots,25\}$$ then we know $$E(18)=24$$ and $$E(19)=5$$, so we have the system

$$\begin{cases} 18a + b & = & 24 &\pmod{26}\\ 19a + b & = &5 &\pmod{26} \end{cases}$$

Substracting the first equation from the second eliminates the $$b$$ and we get $$a = 5-24 = -19 = 7 \pmod{26}$$ and then substituting this back into the first we get $$18\cdot7 + b = 24 \pmod{26}$$, so that $$b=24 - 126 = 2 \pmod{26}$$, and so

$$E(x)=7x+2 \pmod{26}$$

So the decryption function (as the inverse of $$7$$ modulo $$26$$ is $$15$$: $$7 \cdot 15 \equiv 1 \pmod{26}$$ we get that

$$D(x) = 15(x-2) = 15x - 30 = 15x + 22 \pmod{26}$$

which allows us to decrypt the last part of the message WD = $$(22,3)$$ and $$D(22)=14$$, corresponding to O and $$D(3)=15$$ corresponding to P.