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The result below has been disproven.

Let positive integers $p,q,x,y$ satisfy $5p+q = x$ and $5p-q = y$. Prove that if $2y > x$, then $p \geq q$.

Since $2y > x$, we have $2(5p-q) > 5p+q$ and so $5p > 3q$. This result holds true for $(5p+q,5p-q) = (6,4)$, but how do we prove it in general?

2 Answers 2

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It is not true in general, for example take $p=2$ and $q=3$, so $p\not\ge q$

$x=5p+q=13$, $y=5p-q=7$, and $2y>x$

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    Are there any other counterexamples? – 2017-01-04
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    Quite a number of them: (3,4), (4,5), (4,6), (5,6), (5,7), (6, 7)... – 2017-01-04
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    Sorry for being sloppy. I was talking about pairs (p,q), not (x,y). So for p=4 and q=6 you would have x=26 and y=14 – 2017-01-04
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    Generally any positive integers $p$ and $q$ with $p – 2017-01-04
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$$5p+q=x$$ $$5p-q=y$$

Solving:

$$p=\frac{x+y}{10}$$ $$q=\frac{x-y}{2}$$

Then:

$$p-q=\frac{3y-2x}{5}\geq0 \iff y\geq\frac{2}{3}x$$

This result implies that the condition is not in general true.

First obviously: $$2y>x \iff y>\frac{1}{2}x$$ Therefore: $$y\geq\frac{2}{3}x\gt \frac{1}{2}x$$

So if:

$$\frac{1}{2}x \leq y\lt \frac{2}{3}x$$

The hypothesis will hold but the result would not be true.

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    $y\ge\frac{2}{3}x \iff 2y\ge \frac{4}{3}x$, and the hypothesis $2y>x$ does not imply $2y\ge \frac{4}{3}x$, so $p-q\ge 0$ is not necessarily true – 2017-01-04
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    @Momo yes, I said that the condition to be satisfied is more strong because $y\geq \frac{2}{3}x>\frac{1}{2}x$ – 2017-01-04