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I am trying to cryptanalyse a cipher–text encrypted by Affine cipher.

The encryption formula is: $c = f(x) = (ax+b)\bmod m$, where $a$ and $b$ are unknown constants; $x$ is a plain-text symbol, and $c$ the corresponding cipher-text symbol, both in range $[0\dots m-1]$; $m=26$ is known.

This is what I got so far:

$x=4\implies c=17$

$x=19\implies c=10$

That means that letter 4 has been encrypted into 17 (numbers are according to alphabetical order of letters in English language). Same applies to the line below. Now goes the following calculations:

$15a\equiv-7\equiv19\pmod{26}$

$a\equiv19/15\pmod{26}$

$a=3$ [because $15\cdot3=45$ and $45\equiv19\pmod{26}$]

$b=5$

I don’t get the logic which leads to obtaining $b$ value. Any help, please?

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    $\begingroup$ Doesn't your question simply boil down to: "How to get the multiplicative inverse of an element modulo n?" $\endgroup$
    – CodesInChaos
    Commented Aug 18, 2012 at 22:29
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    $\begingroup$ en.wikipedia.org/wiki/Modular_multiplicative_inverse $\endgroup$
    – CodesInChaos
    Commented Aug 18, 2012 at 22:32
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    $\begingroup$ You have $a$, and you have values for $x$ and $c$. Solve for $b$. This question would be more appropriate at Maths SE. $\endgroup$
    – Stephen Harris
    Commented Aug 19, 2012 at 0:15

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Once you know (you didn't question that) that $a = 3$, then knowing 4 goes to 17, using $f(x) = 3x + b$, we just substitute $x = 4$ to get $3 * 4 + b = 12 + b = 17 \mod 26$, where we can easily see that $b= 5$ does the job.

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