Find all solutions $(a, b) \in \mathbb{Z}^{2}$ for each of the following equations

$$
3 a+5 b=1, \quad 3 a+5 b=4, \quad 183 \cdot a+117 \cdot b=3
$$
First, for the first equation, $\left(a_{0}, b_{0}\right)=(2,-1)$ is a solution: $3 \cdot 2+5 \cdot(-1)=1$.

If $(a, b)$ is another solution, then we can write $a=a_{0}+a^{\prime}$ and $b=b_{0}+b^{\prime}$ to obtain

$$
3\left(a_{0}+a^{\prime}\right)+5\left(b_{0}+b^{\prime}\right)=1
$$

so by removing $3 a_{0}+5 b_{0}=1$, we get

$$
3 a^{\prime}+5 b^{\prime}=0.
$$

Therefore, 3 divides $b^{\prime}$ (since 3 divides 0 and 3 and 5 are coprime), i.e., $b^{\prime}=3 k$ for some integer $k$. Then

$$
3 a^{\prime}+15 k=0 \quad \text{so} \quad a^{\prime}=-5 k
$$

The other solution is thus $\left(a_{0}-5 k, b_{0}+3 k\right)=(2-5 k,-1+3 k)$.

It is easy to verify that $(2-5 k,-1+3 k)$ is indeed a solution for all $k$. Therefore, the solutions are all pairs of this form.

For the second equation, since $3 a_{0}+5 b_{0}=1$, we have

$$
3 \cdot\left(4 \cdot a_{0}\right)+5 \cdot\left(4 \cdot b_{0}\right)=4 \cdot 1
$$

So $(8,-4)$ is a solution. We then proceed with the same reasoning as before: if $\left(8+a^{\prime},-4+b^{\prime}\right)$ is another solution, then by simplifying the equation, we get that $b^{\prime}=3 \cdot k$ for some integer $k$ and then that $a^{\prime}=-5 \cdot k$. Therefore, the solutions are exactly the pairs of the form $(8-5 k,-4+3 k)$ for $k$ an integer.

For the third equation, we perform the extended Euclidean algorithm. See the beginner's arithmetic course on the POFM website, page 11: here. We find that $\operatorname{gcd}(183,117)=3$, and we find a solution: $(16,-25)$.

To find all solutions, it is useful to note that the equation is equivalent to

$$
\frac{183}{3} \cdot a+\frac{117}{3} \cdot b=\frac{3}{3}
$$

which is

$$
61 \cdot a+39 \cdot b=1
$$

which can be solved as before to find that the solutions are the pairs of the form

$$
(16-39 \cdot k,-25+61 \cdot k)
$$

for $k \in \mathbb{Z}$.