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Stuck on this homework problem. At this point I know the basic definitions of what it means to be congruent, I just can't seem to link them together to solve this problem, and I'm not really sure how to use the $x+y\equiv 0\ (\mathrm{mod}\ 3)$

I assume I must expand/alter something and then plug it into the $ax+by\equiv 0\ (\mathrm{mod}\ 3)$.

So far for this proof, I know to $a\equiv b\ (\mathrm{mod}\ 3)$, which then implies that $a = b + 3k$, for some $k\in\mathbb{Z}$. Similarly, $b= a + 3j$ for some $j\in\mathbb{Z}$.

Any help on how to link these theorems would be appreciated.

Arctic Char
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vinbera
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1 Answers1

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We can substitute $b=a+3k$ for some $k$ since $a\equiv b\text{ mod }3$

Then $ax+by=ax+(a+3k)y=ax+ay+3yk=a(x+y)+3yk$.

Since $x+y$ is a multiple of $3$, both terms of the sum are divisible by $3$ so $$ax+by\equiv 0\text{ mod }3$$

John Douma
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  • This works, but you shouldn't "unwind" the definition of a congruence like that. Rather, the point of congruences is that they enjoy the same sort of (ring) equational logic that we know well from grade school, so we can effortlessly reuse these equational skills, e.g. as below – Bill Dubuque Nov 12 '20 at 23:56
  • $\qquad\qquad\qquad\begin{align}\color{#c00}a\ \ ,&\equiv, \color{#c00}b\ -x,&\equiv,\ \ y\[.5em] \hline \Rightarrow,\ {-}\color{#c00}ax,&\equiv ,\color{#c00}by\end{align}\ \ \ $ by the Congruence Product Rule – Bill Dubuque Nov 13 '20 at 00:03