Proof of intersection of two positive integer congruence classes I am reviewing the foundation course I took in year 1 while a question caught my eyes：
Let A be the congruence class of 1 mod 3, and B the congruence class of -1 mod 4.
Prove that A∩B is a congruence class mod 12.
The answer is simple, 7 mod 12.
However, I wonder if I need to prove that when m,n happen to be coprime ( hcf(m,n)=1, i.e hcf(3,4)=1 here in particular ), there exist a,b ( a,b belong to integers ) in which am+bn=1 beforehand.
I am not certain about it since the proof of it seems a bit too much for a question phrased as above:
mZ+nZ=gZ for some g which belongs to natural numbers.
g must to be a common factor of m and n, since mZ and nZ are subgroups of gZ.
mZ+nZ (i.e gZ) is contained in every subgroup containing both mZ and nZ, hence,
gZ=mZ+nZ=hcf(m,n)Z
Therefore, 1-(-1)=2=2(4-3)=2*4-2*3,
2*4-1=7=2*3+1
lcm(3,4)=12
May I ask do I really need to write down the proof of existence of (a,b) such that am+bn=1 when hcf(m,n)=1in order to answer this question? 
Also, I really struggle to explain how I come up with 12 here.
Thank you so much！
Regards,
 A: If a number can be represented as both $3k+1$ and $4l-1$, then that number can also be represented as $12m+7$. This can be seen by a simple substitution:
Say $x=3k+1 = 4l-1$
$3k +2 = 4l$
$9k+6=12l$
$k+6 = 4(3l-2k) $
$k-2 = 4(3l-2k-2)$
$k = 2+4m$ 
Substituting $k$ back in $x$ gives
$$x=3k+1=3(4m+2)+1 = 12m+7$$
A: Hint $ $ if $\,x_0 \in A\cap B\,$ then $\,x \in A\cap B \!\iff\!  3,4\mid x\!-\!x_0\!\color{#c00}\iff\! 12\mid x\!-\!x_0\!\iff\! \bbox[6px,border:1px solid #c00]{x \in x_0\!+\!12\Bbb Z}$
Remark $ $ Generally for moduli $m,n\!:\ m,n\mid x\!-\!x_0\color{#c00}\iff {\rm lcm}(m,n)\mid x\!-\! x_0.\,$ OP is the special case where $\,\gcd(m,n) = 1\  [\!\iff {\rm lcm}(m,n) = mn\:\!$].
This is equivalent to the uniqueness half or CRT = Chinese Remainder Theorem (the other half = existence states that such an $\,x_0\,$  exsists). The view in terms of cosets becomes clearer if you study the ring-theoretic form of CRT, i.e. 
$$ \gcd(m,n)=1\ \Rightarrow\ \Bbb Z/m \times \Bbb Z/n\, \cong\, \Bbb Z/mn\qquad\qquad\qquad$$ 
A: Every congruence class modulo $3$ is the union of four congruence classes modulo $12$. Every congruence class modulo $4$ is the union of three congruence classes modulo $12$. So $A\cap B$ is the union of some number of congruence classes modulo $12$. To prove it's precisely one class, just try all $12$ classes and see that elements of $11$ of them don't work. 
