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Prove that $xRy \iff 5 |(x+4y)$ is an equivalence relation.

Reflexive: $xRx$, since $x+4x = 5x$, which is a multiple of $5$.

Transitive: Suppose $xRy$ and $yRz$. Then $$x+4y=5k_1,\quad y+4z = 5k_2.$$ After adding the two and some algebraic manipulation we obtain $$x+4z = 5(k_1+k_2-y)$$ thus $xRz$.

Symmetric: Suppose $xRy$. I need to show that $yRx$. I am having trouble showing this, however. Can anyone please help me?

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    $\begingroup$ Hint: What is $(x+4y)+(y+4x)$? $\endgroup$ – Slade Nov 17 '15 at 22:55
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Note that $$ x+4y = x + (5-1)y = 5y + (x-y), $$ so $xRy \iff 5|(x-y)$. I assume you're working in $\Bbb Z$, all integers — positive, negative and $0$. If $n|m$ then $n|(-m)$, so if $5|(x-y)$ then $5|(y-x)$.

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We will assume $xRy$. So, this means that there is some integer $k$ such that $x+4y = 5k$. This implies that we can write $x = 5k - 4y$.

This means that $4x + y = 4(5k - 4y) + y = 20k - 15y = 5(4k-3y).$

Hence $4x+y = 5(4k-3y),$ which is clearly divisible by $5.$ We conclude $yRx.$

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