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I was solving this question, but I'm hitting a wall.

Find all pairs of $(x,y)$ such that $(2x+7y)|(7x+2y)$

Here is what I have done.

Since we have two variables, this question has either a small number of solutions, or all solutions are of the same form. I assume the first.

So, $(2x+7y)|(98x+14y)$, and then $(4x+14y)|(98x+14y)$. Then $4x|98x$.

Which I can't find any examples of. Is there something wrong with my approach? I can't seem to find the solutions.

Can anyone help?

Community
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DynamoBlaze
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  • Are $x$ and $y$ integers? Real numbers? In any case, use the definition of divisibility and start with the thought that $2x+7y=k\cdot (7x+2y)$ for some integer $k$. Then, treat it like a linear equation. – Harry Jul 15 '17 at 04:03
  • @Harry They are integers. But a single equation and three(two) variables? How do I do so? – DynamoBlaze Jul 15 '17 at 04:05
  • That equation can be manipulated until we have an explicit expression for $x$ or $y$. From earlier, we have that $x\cdot (2-7k)+y\cdot (7-2k)=0$. Try isolating $x$ and seeing what kinds of $y$ satisfy the equality. – Harry Jul 15 '17 at 04:11

1 Answers1

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I will assume all positive integer numbers.

$x=4y$ and $x=19y$ that is $(4t;\;t)$ and $(19u;\;u)$ for any $t,\;u\in \mathbb{N}-\{0\}$

Indeed $(2x+7y)|(7x+2y)$ if there exist $k\in\mathbb{N}-\{0\}$ such that $$(7 x + 2 y) = k (2 x + 7 y)$$ which gives $$x= \frac{y(2 -7 k )}{2 k-7}$$ That is integer only if $\dfrac{2-7 k}{2 k-7}$ is integer which happens for $k=1$ that is not good because $x$ is negative, $k=2$ which gives $x=4y$ and $k=3$ which gives $x=19y$

If you want negative solutions, too, you can do in a similar way

Hope this helps

Raffaele
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