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For three integers $p, q, r$ such that $\gcd(p, q)$ $=$ $1$, $\gcd(r, q)$ $=$ $1$, let $d = pr$.

$(p+q)(q+d) - pq = x$

$(q+d)q - p(p+q) = y$

Prove that

$rx-q-d$ = $y$

Thanks if someone can at least simplify the proof to three terms instead of six.

1 Answers 1

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The proposition does not hold true. Counterexample: $p=2, q=3, r=5$ so that $d=2 \cdot 5 = 10\,$:

$$ \begin{align} x & = 5 \cdot 13 - 2 \cdot 3 = 59 \\ y & = 13 \cdot 3 - 2 \cdot 5 = 29 \\ rx - q - d & = 5\cdot59-3-10=282 \;\;\ne\;\; 29 = y \end{align} $$

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    The proposition is not always true in general, it seems to hold true when the three parameters are of the form $p$, $q = pr+1$, $r$. The polynomial extension(s) are very complicated though. – 2017-02-03
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    @J.Linne It is possible that the proposition may hold true with additional constraints, but that's a different question. – 2017-02-03
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    @J.Linne FWIW it doesn't hold true for $p=1, r=2, q=1\cdot 2+1 = 3, d=1 \cdot 2 = 2\,$. – 2017-02-03