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I'm having trouble solving the following congruence:

$$2x + 2y ≡ 0 \pmod{7}.$$

What I've tried so far is writing the congruence as follows: since the remainder is $0,$ we know $7$ is divisible by $2x + 2y,$ so $2x + 2y = 0 + 7k,$ and I got pretty much stuck on this step, so any further help is greatly appreciated.

J. W. Tanner
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kokayy
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  • Since $\gcd(2,7)=1$, so from this you can further conclude that $x+y \equiv 0 \pmod{7}$. Without any other information about $x$ and $y$ this is the best you can conclude. – Anurag A Jun 18 '19 at 19:31
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    $7|2x+2y=2(x+y)\implies7|2$ or $7|x+y$. But $7\not|2$, so this means $7|x+y.$ Conversely, $7|x+y\implies7|2x+2y$. – J. W. Tanner Jun 18 '19 at 19:38

2 Answers2

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Multiplying both sides by $4$, $$8x+8y\equiv x+y\equiv0\mod{7}$$ Hence $$y\equiv -x\mod{7}$$ $$\therefore y=-x+7k$$ where $k\in\mathbb{Z}$.

Peter Foreman
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$7 | 2x+2y =2(x+y) \implies 7|2 $ or $7|x+y$. But $7\not|2$, so this means $7|x+y$.

Conversely, $7|x+y\implies7|2x+2y$.

J. W. Tanner
  • 60,406