Find a solution using a Diophantine equation The diophantine equation is 3a + 12b = 132
From the textbook, I set out to find the gcd(3,12) which is 3. 
then I proceed to set up the equation 3s+12t = 0 for any integer s and t
0 = 12(1) + 3(-4)
0 = 12(132) + 3(-4*132)
0 = 12(132 +3m) + 3(-528 -12m)
solve for m 
132 +3m =< 0
3m =< -132
m =< -44
but now I have no clue how to proceed from here.
Thanks in advance for your help 
Note: I am trying to find the positive solutions

The greatest common divisor of the coefficients of the Diophantine equation $3a + 12b = 132$ is $\gcd(3, 12, 132) = 3$.  Divide by $3$ to obtain 
$$a + 4b = 44$$
A particular solution in the positive integers is $(1, 10)$.  If $b$ decreases by $1$, then $a$ must increase by $4$.  Hence, the solutions in the positive integers are of the form
\begin{align*}
a & = 4t\\
b & = 11 - t
\end{align*}
where $t \in \mathbb{Z}$, $1 \leq t \leq 10$.  Consequently, the possible solutions in the positive integers are 
$$(4, 10), (8, 9), (12, 8), (16, 7), (20, 6), (24, 5), (28, 4), (32, 3), (36, 2), (40, 1)$$
