Solving Linear Congruences over Cryptography I have got some doubt in cryptography mainly related to Linear Congruences.

Question

Suppose that the most common letter and the second most common letter in a long 
ciphertext produced by encrypting a plaintext using an Afﬁne Cipher 
$f\left(p \right)=\left (ap + b \right)\,\pmod{26} $are  $Z$ and $J$ respectively.
 What are the most likely values of a and b?
Converting Alphabets into Numerals we have $Z=25,J=9$
I was stuck. I looked into solution.In the they have assumed
$f\left(4 \right)=25$ and $f\left(19 \right)=9$
I am not getting how they assuming  to take $p=4$  for $f\left(p \right)=25$
and $p=19$  for $f\left(p \right)=9$. I closed the solution and solved the problem by taking

$f\left(4 \right)=25$ and $f\left(19 \right)=9$ 
My Solution

Affine Cipher is of the form 
$f\left(p \right) \equiv \left(ap+b \right)\,\pmod{26}$
$f\left(4 \right)=25$ and $f\left(19 \right)=9$ 
Equations-:


*

*$\Rightarrow 25 \equiv \left(4a+b \right)\,\pmod{26}$ 

*$\Rightarrow 9 \equiv \left(19a+b \right)\,\pmod{26}$


Subtracting 1 from 2,
$\Rightarrow 10 \equiv \left(15a \right)\,\pmod{26}$
$\Rightarrow 2 \equiv \left(3a \right)\,\pmod{26}$
$\Rightarrow a \equiv 2*3^{-1}\,\pmod{26}$
$\Rightarrow 3^{-1}\Rightarrow$ inverse of $3\,\pmod{26}=9$
$\Rightarrow a \equiv 18\,\pmod{26}=18$
Now,
$\Rightarrow 4a+b \equiv 25\,\pmod{26}$
$\Rightarrow 4*18+b \equiv 25\,\pmod{26}$
$\Rightarrow b \equiv 5\,\pmod{26}$
Answer is Correct.But Only thing i am not getting 
How $f\left(4 \right)=25$ and $f\left(19 \right)=9$ ???

I don't know why you're not getting $25$ and $19$, because that's what I get:
$$f(4) = 18\cdot 4 + 5 = 77 \equiv 25 \pmod{26}.$$
And 
$$f(19) = 18\cdot 19 +5 = 347 = 9 \pmod{26}.$$
