# Determine values of $\ p , \ q \$ if we have$\ (3 \mathbb{Z}+1) \cap (4 \mathbb{Z}+2) =(p \mathbb{Z}+q) \$

Determine values of $\ p , \ q \$ if we have $\ (3 \mathbb{Z}+1) \cap (4 \mathbb{Z}+2) =(p \mathbb{Z}+q) \$

Also use set comprehension notation for $\ (r \mathbb{Z}+t) \cup 3 \mathbb{Z} \ , \ \ r \neq t , \ \ r,t \in \mathbb{Z}$

Answer: $(3 \mathbb{Z}+1) \cap (4 \mathbb{Z}+2) =(p \mathbb{Z}+q) \\ \Rightarrow \{a \in \mathbb{Z} | a=3k+1 \} \cap \{b \in \mathbb{Z} | b=4k'+2 \}=(p \mathbb{Z}+q) \\ \Rightarrow \{a \in \mathbb{Z} | a=b \} =(p \mathbb{Z}+q)$

But I can't proceed further .

If there is any help ?

You can use the Chinese Remainder Theorem. Since $3$ and $4$ are coprime, the set of equations: $$x = 1 \mod 3 \\ x = 2 \mod 4$$
has a solution which is unique modulo $3\cdot4 = 12$
By observing we can find that the solution is $x = 10$, and since it is unique modulo $12$ we can deduce that $(3 \mathbb{Z}+1) \cap (4 \mathbb{Z}+2) =(12 \mathbb{Z}+10)$
Without observing, there's a formula to solve a system of two linear congruences modulo coprime integers $a, b$, which relies on a Bézout's identity between these integers: $$ua+vb=1.$$ Such a relation can always be found with the extended Euclidean algorithm (which is not necessary here: $\;4-3=1$). $$\begin{cases}x\equiv r\pmod a\\x\equiv s\pmod b\end{cases}\iff x\equiv sua+rvb\pmod{ab}.$$ In the present case, you obtain $\;\color{red}{x\equiv -2 \mod{12}}$.
$3\Bbb Z+1 = \{...,-2, 1,4,7,10,13,16,19,22,...\} \\ 4\Bbb Z+2 = \{...,-2,2,6,10,14,18,22,...\}\\3\Bbb Z+1\cap4\Bbb Z+2=\{...,-2,10,22,...\}$