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Given two letters of the plaintext 'ZP' corresponding to the ciphertext 'AE' respectively, I have found the key to be (a,b)=(10,10) (modulo 26).

Therefore the decryption map $D(x)=a^{-1}(x-a)\bmod26$ I obtain here is $D(x)=10^{-1}(x-10) \bmod26$. However, $10^{-1}\bmod26$ does not exist so how am I able to find this? Have I got the key wrong? I know $(23,23)$ also works but i dont know how to find that key with my working.

This is my working:

$25a+b\equiv0 \pmod{26}$, $15a+b\equiv4 \pmod{26}$

where $(a,b)$ are the two elements of the key, and $Z,A,P,E$ correspond to the numbers $25,0,15,4$ using the standard translation of the alphabet to numbers, starting with $A=0$.

Then using simultaneous equations I get:

$10a\equiv-4\pmod{26}$ which can be written as $10a\equiv22\pmod{26}$ as $-4\equiv22\pmod{26}$.

Then dividing by $2$ you get $5a\equiv11\pmod{13}$ so $a\equiv11\cdot5^{-1}\pmod{13}$ then by using the extended euclidean algorithm you can calculate the inverse of $5\bmod 26$ which is $-5\pmod{13}$ or $8\pmod{13}$.

Therefore $a\equiv11\cdot 8 \pmod{13}$, $a\equiv88\pmod{13}$ and so you get $a\equiv10\pmod{13}$.

Therefore $a=10$ and by plugging $a=10$ into $25a+b\equiv0\pmod{26}$,

I get $250+b\equiv0\pmod{26}$ and so $b\equiv-250\pmod{26}$ which then gives $b=10$ and so $(a,b)=(10,10)$.

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