Prove: A ⋂ (B+C) = (A ⋂ B) + (A ⋂ C) Please, help me to prove this:
For $A,B,C$ subgroups in $\mathbb{Z}$:
$A\cap (B+C) = (A \cap  B) + (A \cap  C)$
All that I know about this solution is, with
$A=a\mathbb{Z},\ B=b\mathbb{Z},\ C=c\mathbb{Z}$, 
$A + B = u\mathbb{Z}$, and $u = gcd(a,b)$
$A \cap  B = dZ$, and $d = lcm(a,b)$
I tried to rewrite this like $lcm(a,gcd(b,c)) = gcd(lcm(a,b),lcm(a,c))$, and like composition of prime powers but it didn't help me.
 A: As $x\mathbb{Z}=-x\mathbb{Z}$, I can suppose that all generators $a,b,c$ are positive. 
If one of them is zero, the relation is straightforward (you can write it completely as an exercise). 
Now, inside $\mathbb{N}_{\geq 1}=\{1,2,\cdots\}$, divisibility relation (write it with a vertical bar) is an order relation with 
$$
lcm=max_| \mbox{ and } gcd=min_|
$$
a golden rule for any ordering $\preceq$ (with $max$ and $min$ i.e. a lattice order) is 
$$
X\preceq min(Y,Z) \Longleftrightarrow X\preceq Y \mbox{ and }  X\preceq Z\qquad (R1)
$$ 
and 
$$
max(Y,Z)\preceq X \Longleftrightarrow Y\preceq X \mbox{ and }  Z\preceq X\qquad (R2)
$$ 
this helps decompose the problem into subproblems.
Now to prove $lcm(a,gcd(b,c)) = gcd(lcm(a,b),lcm(a,c))$, you just have 
to decompose this in two parts $lcm(a,gcd(b,c)) \preceq gcd(lcm(a,b),lcm(a,c))$ and 
$lcm(a,gcd(b,c)) \succeq gcd(lcm(a,b),lcm(a,c))$,
($X\preceq Y$ being your divisibility relation $X|Y$) and use (R1-R2). 
Another route As you said at the end "composition of prime powers", I indicate you this route as well.
I still suppose that all generators $a,b,c$ are positive and non-zero.
Every integer $m>0$ can be written uniquely ($\mathfrak{p}$ is the set of prime numbers)
$$
m=\prod_{p\in \mathfrak{p}}p^{\nu(p,m)}
$$ 
For every $m$, let us note $\nu(?,m)$ the function $p\mapsto \nu(p,m)$. It is easily seen that 
$$
\nu(?,lcm(x,y))=sup(\nu(?,x),\nu(?,y))\mbox{ and }
\nu(?,gcd(x,y))=inf(\nu(?,x),\nu(?,y)) 
$$
so, the identity $lcm(a,gcd(b,c)) = gcd(lcm(a,b),lcm(a,c))$ amonts to show $\nu(?,LHS)=\nu(?,RHS)$ i.e, for every prime $p\in \mathfrak{p}$
$$
sup\Big(\nu(p,a),inf(\nu(p,b),\nu(p,c))\Big)=
inf\Big(sup(\nu(p,a),\nu(p,b)),sup(\nu(p,a),\nu(p,c))\Big)
$$
which is straightforward.
Hope it helps.
Late edit One can use, equivalently, double inclusion for $A\cap(B+C)= (A\cap B)+(A\cap C)$. One move (i.e. $\supset$) is easy and uses only elementary arguments, but the other one (i.e. $A\cap(B+C)\subset A\cap B+A\cap C$) is specific to $\mathbb{Z}$ as it does not hold for other groups. Take, for example, $G=\mathbb{Z}^2,\ A=(1,1).G,\ B=(1,0).G,\ C=(0,1).G$, one has $A\cap(B+C)=A$ whereas $A\cap B=A\cap C=\{(0,0)\}$.  
_
A: 1) First show that $A\cap(B+C)\subset A\cap B+A\cap C$.
Let $x\in A\cap(B+C)$, then $x\in A$ and $x\in (B+C)$. Also let $x=z+y$ such that $y\in B$ and $z\in C$, then $y\in A\cap B$ and $z\in A\cap C$, thus $x=y+z\in A\cap B+A\cap C$.
2) Now show that $A\cap(B+C)\supset A\cap B+A\cap C$.
Let $x\in A\cap B$ and $y\in A\cap C$, then $x\in A$, $x\in B$ and $y\in A$, $y\in C$. Then we have $x+y\in A$ and $x+y\in B+C$, which means $x+y\in A\cap(B+C)$.
