How to prove $ (3\Bbb N + 1) \cap (5\Bbb N + 3) = 15\Bbb N + 13$

Question

I'm trying to prove that: $$ (3\Bbb N + 1) \cap (5\Bbb N + 3) = 15\Bbb N + 13$$ What have I done so far is, I'm getting an x from left hand side of equation: $$ x \in (3\Bbb N + 1) \cap (5\Bbb N + 3) $$ Then I'm simplifying it, like this: $$ x \in (3\Bbb N + 1) \land x \in (5\Bbb N + 3) $$ $3\Bbb N + 1$ is basicly $1 \pmod{3}$, and $(5\Bbb N + 3)$ is basicly $3 \pmod{5}$, thus: $$ x \equiv 1 \pmod{3} \; and \; x\equiv 3 \pmod{5} $$ Use CRT (i.e. https://en.wikipedia.org/wiki/Chinese_remainder_theorem) to conclude result for left hand side: $$ a_1 = 1, \; m_1 = 5, \; y_1 \equiv 2 \pmod{3} $$ After that, Use CRT to conclude result for right hand side: $$ a_2 = 3, \; m_2 = 3, \; y_2 = 2 \pmod{5} $$ Then plug in: $$ \begin{split} x &\equiv a_1m_1y_1 + a_2m_2y_2 \pmod{M} \\ & \equiv 1\cdot 5\cdot 2 + 3\cdot 3 \cdot 2 \pmod{3\cdot 5} \\ & \equiv 28 \pmod{15} \equiv 13 \pmod{15} \end{split} $$ then find for the right side; $$ x \in (15\Bbb N + 13) \implies x = 13 \pmod{15} $$ Did I prove it right? Is there an easier way to do this?

Answer 1

Let $n=3x+1$ and assume $x=5y+z$, then $n=15y+3z+1$. If $n\equiv 3 \pmod{5}$, then $z\equiv -1 \pmod5$. Thus we can take $z=-1$ and obtain the result you want.

Answer 2

Consider the ring isomorphism $\phi:\Bbb Z_{15}\rightarrow \Bbb Z_3\times \Bbb Z_5: x\mapsto (x \mod 3,x\mod 5)$. Then $x=13$ is mapped to $\phi(13) = (13\mod 3,13\mod 5)= (1,3)$.