I'm currently learning Diophantine equations, and does not get the exact same answer as the course literature, I would love some help to understand if my answer corresponds to the books answer.
$\textbf{Question 1}$
$\bullet$ Solve $8x+9y=3$
$$GCD(8,9)=1$$
$$1|3 \Rightarrow \text {solution exists}$$
Euclides algoritm backwards gives me that $$\begin{cases} x=3\\ y=-3 \end{cases}$$
So far so good, now comes the part that doesn't seem right.
All solution according to me is given by
$$\begin{cases} x=3+\Big(\frac{9}{1}\Big)n=3+9n\\ y=-3-\Big(\frac{8}{1}\Big)n=-3-8n \end{cases}$$
Where as the book says $$\begin{cases} x=-(3+9n) \\ y=3+8n \end{cases}$$
Does the sign matters, or is this basically the same answer?
$\textbf{Question 2}$
$\bullet$ Solve 36x+78y=18$
$$GCD(36,78)=6$$
$$6|18 \Rightarrow \text {solution exists}$$
Euclides algoritm backwards gives me that $$\begin{cases} x=-6\\ y=3 \end{cases}$$
So far so good again, now comes the part that doesn't seem right.
All solution according to me is given by
$$\begin{cases} x=-6 + \Big(\frac{78}{6}\Big)n=-6+13n\\ y=3-\Big(\frac{36}{6}\Big)n=3-6n \end{cases}$$
Where as the book says $$\begin{cases} x=-(6+13n) \\ y=3+6n \end{cases}$$
Does the sign matters, or is this basically the same answer?
For the first question, $x = 3$, $y = -3$ is not a solution. Indeed, $8\cdot 3 + 9 \cdot(-3) = -3$.
In the second question the answers are equivalent and to explain why, let's look at what $x = -6+13n$, $y = 3-6n$ means.
First thing to mention is that you forgot to write the key part and that is $n\in\mathbb Z$. Then, $x = -6+13n$, $y = 3-6n$, $n\in\mathbb Z$, is just a way to write that the set of integer solutions is $\{(-6+13n,3-6n)\mid n\in\mathbb Z\}$. But, this is the same as the set $\{(-6-13k,3+6k)\mid k\in\mathbb Z\}$, they are not just in bijection, they are equal sets.
To see why, take a member of the first set, $(-6+13n,3-6n)$ for some $n\in\mathbb Z$ and set $k = -n$. Then, $(-6+13n,3-6n) = (-6-13k,3+6k)$ and $k\in\mathbb Z$, so it's also in the second set. You can show that any member of the second set is in the first set analogously.
To make sure that what you wrote is correct, you can simply substitute into the starting equation: $$36(-6+13n)+78(3-6n) = 36\cdot(-6) + 78\cdot 3 + 36\cdot 13n - 78\cdot 6n = 18.$$