How to combine congruences? I have two congruences:
$$ \text{(i) }p\equiv 1 \mod 3 \,\,\, \land \,\,\, p\equiv 1 \mod 4 \\
\text{(ii) }p\equiv 2 \mod 3 \,\,\, \land \,\,\, p\equiv 3 \mod 4
$$
Is it possible to write these systems of congruences into a single congruence? To be clear: I don't want to combine (i) and (ii), but want to merge each statement into a single expression. 
 A: Yes. Since $3$ and $4$ are coprime, you need the Chinese remainder theorem for that. Namely

*

*(i) is equivalent to $x\equiv 1\mod 12$.

*(ii) requires the explicit formulation of the inverse isomorphism from $\mathbf Z/3\mathbf Z\times \mathbf Z/4\mathbf Z \to \mathbf Z/12\mathbf Z$. Start from a Bézout's relation between $3$ and $4$: $4-3=1$. The solution is given by
$$x\equiv  \color{red}2\cdot\color{lightgreen} 4-\color{lightgreen}3\cdot \color{red}3=-1\equiv 11\mod 12.$$
Added: you may write the general formula for the system of congruences modulo coprime numbers, given a Bézout's relation between $a$ and $b$:  $ua+vb=1\quad(u,v\in \mathbf Z)$, $$\begin{cases} x\equiv \color{red}\alpha\mod \color{red}a \\ x\equiv \color{lightgreen}\beta\mod \color{lightgreen}b \end{cases} \iff x\equiv \color{lightgreen}\beta\,u\color{red}{a}+\color{red}\alpha\,v\color{lightgreen}{b}\mod ab.$$
A: Since $\gcd(3,4) = 1$, by the Chinese Remainder Theorem, the solution is given by


*

*$p \equiv 1 \pmod{12}$

*$p \equiv 11 \pmod{12}$

