It seems like it would be wise for you to brush up on the idea of what modular arithmetic is and what it really means. It's true that x = -1 would be the only solution to the equation \(4x + 5 = 1\), and likewise x = 6 would be the only solution to the equation \(4x - 24 = 0\) if you were working under the "normal" rules of arithmetic, but that's *not* what's being asked here.

The notation \(\displaystyle \frac{\mathbb{Z}}{N\mathbb{Z}}\) means you're working in a ring where you only have access to the integers \(0, 1, 2, \cdots, N-1\) and when you would reach \(N\), you "roll over" and begin again at 0. So, if you were doing some workings and you found an answer of 27 but you were working in \(\displaystyle \frac{\mathbb{Z}}{8\mathbb{Z}}\), you'd have to find what element 27 is congruent to... 27 - 8 = 19, 19 - 8 = 11, 11 - 8 = 3, so we say that \(27 \equiv 3 \: (\text{mod} \: 8)\)

As a full example of this concept, consider finding all solutions to \(3x + 2 = 2\) in \(\displaystyle \frac{\mathbb{Z}}{6\mathbb{Z}}\). As a visual means of further demonstrating the idea, let's manually substitute in each value \(0, 1, 2, \cdots, 5\), and see what we get:

\(\displaystyle 3(0) + 2 \equiv 2 \: (\text{mod} \: 6)\)

\(\displaystyle 3(1) + 2 \equiv 5 \: (\text{mod} \: 6)\)

\(\displaystyle 3(2) + 2 \equiv 2 \: (\text{mod} \: 6) \: \: \: {\color{red} \textit{3(2) + 2 = 8, but 8 is not an element of the ring, so we "reduce" it to one that is}}\)

\(\displaystyle 3(3) + 2 \equiv 5 \: (\text{mod} \: 6) \: \: \: {\color{red} \textit{3(3) + 2 = 11, but 11 is not an element of the ring, so we "reduce" it to one that is}}\)

\(\displaystyle 3(4) + 2 \equiv 2 \: (\text{mod} \: 6)\)

\(\displaystyle 3(5) + 2 \equiv 5 \: (\text{mod} \: 6)\)

Now you give it a try. It happens to be true in both cases that the solution under "normal" arithmetic is also a solution in the new ring, but are there other solutions to these modular equations? What might they be?