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In order to solve the congruence 2x + 6 ≡ 4 (mod 8), your friend PhilLovett 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 to2x + 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.
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