The encoding formula is $y=3x+21$ and it arrived at a decoding formula of $x=9y+19*$ (using mod 26).
I'm just confused on how they got the decoding formula/the inverse of the original equation? or how the x was isolated?
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1$3^{-1} = 9 \bmod 26$, can you find the rest? – kelalaka Mar 16 '22 at 18:47
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$9y = 27x + 189 \implies 9y - 189 = 27x.$ – user2661923 Mar 16 '22 at 18:58
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$!\bmod 26!:\ y \equiv 3x!+!21 \iff 3x \equiv y!-!21\iff x \equiv y\color{#c00}{/3} - 7\equiv \color{#c00}9y-7, $ by $,\color{#c00}{1/3}\equiv 27/3\equiv \color{#c00}9,,$ as explained in the linked dupe. – Bill Dubuque Mar 16 '22 at 21:36
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Here is a simple way to show it:
Since $y=(3x+21) \mod 26$ you can equivalently write using identities of the remainders
$$3x=(y-21) \mod 26\Rightarrow 27x=(9y+19) \mod 26 $$
however since $27x=x+26x=x\mod 26$ the result follows.
More generally, to solve the equation $$px+qy=r\mod n,~ \gcd(p,n)=\gcd(q,n)=1$$
for $x$ or $y$ respectively, you need to find numbers $(a,b)$ such that
$$ap+bn=1$$
which is possible according to Bezout's identity, and one pair can always be found using the extended Euclidean algorithm. In our case, $p=3, n=26$ and the pair returned by the Euclidean algorithm is $a=9,b=-1$.
DinosaurEgg
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Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 16 '22 at 21:31
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@BillDubuque in this case we merge the questions in cryptography.se. Isn't this a good practice? – kelalaka Mar 18 '22 at 19:33