If there exists a parallelogram (not a rectangle) $ABCD$ such that $AB=CD=x$, $BC=DA=y$, $AC=z$, $BD=w$, do there exist postive distinct integers $x,y,z,w$ such that $x\geq y\geq z\geq w$?
I know we have $$2(x^2+y^2)=z^2+w^2$$
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lioness99a
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math110
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if x,y,z,w are distinct, the parallelogram can't be a rectangle anyway. – Jun 14 '17 at 14:54
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Can we have $x=y$? – Asinomás Jun 14 '17 at 14:55
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$x\ge y\ge z\ge w$ – math110 Jun 14 '17 at 14:55
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1$x\geq y\geq z\geq w$ clearly can't be true, since then $2(x^2+y^2)>x^2+y^2\geq z^2+w^2$. – Wojowu Jun 14 '17 at 14:59
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must the vertices be labeled in that order? is the paralelogram convex? – Asinomás Jun 14 '17 at 14:59
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2https://math.stackexchange.com/questions/1127654/parametrization-of-solutions-of-diophantine-equation – individ Jun 14 '17 at 15:00
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2@JorgeFernándezHidalgo All parallelograms are convex. – Harald Hanche-Olsen Jun 14 '17 at 15:00
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Possibly related: https://math.stackexchange.com/questions/1254318/combinatorics-integer-sided-triangles-with-integer-median – user49640 Jun 14 '17 at 15:00
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@individ That equation is slightly different, lacking the factor $2$. – Harald Hanche-Olsen Jun 14 '17 at 15:01
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@HaraldHanche-Olsen same equation. Put the right coefficients. – individ Jun 14 '17 at 15:03
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@HaraldHanche-Olsen If $x^2+y^2=z^2+w^2$ then $(z+w)^2+(z-w)^2=2(x^2+y^2)$ (an application of the Brahmagupta identity, really). Therefore solutions of the former give solutions of the latter. – Jyrki Lahtonen Jun 14 '17 at 15:05
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2Mustn't one of $z$ and $w$ be larger than both $x$ and $y$, contradicting $x\geq y\geq z\geq w$? – md2perpe Jun 14 '17 at 15:12
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@md2perpe I missed that (it was only a comment when I posted). – Jyrki Lahtonen Jun 22 '17 at 22:59
2 Answers
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An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no prime congruent to $3 (\mod 4)$ raised to an odd power. Thus there are such infinitely many integers without $x\geq y\geq z\geq w$. Even more simply $2(x^2+y^2)=(x+y)^2+(x-y)^2.$
But with the condition $x\geq y\geq z\geq w,$ no such quadruples.
Bumblebee
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For $x\leq y\leq z\leq w$ there is the following $x=68$, $y=85$, $z=87$ and $w=127$
Michael Rozenberg
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