In order to solve the congruence 2x + 6 ≡ 4 (mod 8), your friend Phil
Lovett wrote down the following steps:

2x+6 ≡ 4 (mod 8)
x+3 ≡ 2 (mod8)
x ≡ -1 (mod 8)

From here, Phil concludes that the solution set to
2x + 6 ≡ 4 (mod 8) is {x; x ≡ -1 (mod 8)}.

(a) Is Phil's answer correct? If not, which step in Phil's reasoning is incorrect? Explain what is wrong with it and find the (correct) solutions for the equation 2x+6 ≡ 4 (mod 8) .

(b) Find the condition for the modulus n, for which the congruence  2x+6 ≡ 4 (mod 8) actually does have the solution set {x; x ≡ -1 (mod 8)}. (i.e. it is the set of all solutions). Prove your answer.