# How to show $5~|~y$ if $y = x - 4$?

So, I have to find the solution to the congruencies

$$x \equiv 4 \mod 315$$

$$x \equiv 9 \mod 715$$

I know that I need to divide through by 5 to make the moduli coprime, and I know how to solve for $$x$$ from there. What I'm having trouble with is figuring out how to divide through by 5. I was given the following hint:

Hint: Let $$y = x−4$$, show that y must be divisible by $$5$$.

However, I'm not sure how I should show that $$5 ~|~ x−4$$.

Any help would be appreciated.

• From $5\mid315$, it follows that, if $315\mid x-4$, then $5\mid x-4$. Do you know the Chinese remainder theorem? May 24 at 1:45
• May 24 at 1:53
• @J.W.Tanner I do know the Chinese Remainder Theorem, I just wasn't sure how to get $5|x-4$ but clearly I was overthinking things. Thanks for your help. May 24 at 2:03
• "However, I'm not sure how I should show that 5|x−4.". Well....$315| x-4$ so $5|x-4$ (and $63|x-4$ but that's not relevant). Likewise $715|x-9$ so $5|x-9$ so $5|x-9+5 = x-4$ (and $143|x-9$ and so $143|x-9+143=x+134$ but that's not relevant either) May 24 at 3:12
• A good rule is if $a \equiv b \mod n$ and $m|n$ then $a\equiv b \pmod m$ as well. Do you see why?[1] Conversely. If $c\equiv d \pmod m$ then $c \equiv d + km$ for some integer $k: 0 \le k < \frac nm$. [1] The reason why is $a\equiv b \pmod n \iff n| a-b$ but if $m|n$ and $n|a-b$ then $m|a-b$. May 24 at 3:15

I'm not sure how I should show that $$5 \mid x−4$$.
You have $$x\equiv 4\bmod 315$$. That means $$315\mid x-4$$.
Since $$5\mid315$$, it follows from the transitive property of divisibility that $$5\mid x-4$$.
We know that $$x \cong 4 \mod 315$$, and therefore $$x-4 \cong 4-4 \mod 315$$ (because addition and subtraction preserve congruence).
$$y \cong 0 \mod 315$$ is equivalent to $$315 \mid y$$, which is equivalent to $$y = 315 \times p$$ for some natural $$p$$. If we break 315 down into prime factors, we get that $$y = 3 \times 3 \times 5 \times 7 \times p$$, and thus $$5 \mid y$$.