How many oracle calls to break an affine cipher in a chosen plaintext attack? Assuming the ability to launch Chosen Plaintext Attacks (CPA), how many oracle calls does an attacker need to break the affine cipher? How does it work?
The answer is 1 I think because if we apply a chosen plain text attack, then we need at least 2 encrypted alphabets and two plaintext alphabets corresponding to the related encrypted alphabets. Then we will have two variable and two equations. 

By solving those you can get the additive key as well as the multiplicative key.