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This is "Question 989" of S. Malhari Rao submitted to JIMS vol. XIV, 118 .

A solution is:

$A=\kappa^{4}+\kappa^{2}+1$

$$x=\frac{A^{2}}{\kappa^{6}}$$

$$y=\frac{A^{2}}{\kappa^{2}}$$

$$z=\frac{A^{2}}{\kappa^{4}}$$

Very different from that given by K. J. Sanjana and N. B. Mitra.

Can someone find the relationship with whole solutions?

giuseppe mancò
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2 Answers2

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The Diophantine equation $$ (x+y+z)^2=xyz $$ can be solved by using Pell's equation, see here. The given rational solution is certainly integral for $\kappa=\pm 1$, but then only gives $x=y=z=9$. This is the obvious integral solution for $x=y=z$, besides $x=y=z=0$.

Dietrich Burde
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  • Hi, Dietrich. You and I both answered this (allowing a coefficient) years ago https://math.stackexchange.com/questions/1930438/equation-with-vieta-jumping-xyz2-nxyz I note the question says rational but the OP seems satisfied with a list of integer solutions; – Will Jagy May 29 '20 at 19:56
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    Hi, Will. You are right. I had totally forgotten this. You have a great answer there. – Dietrich Burde May 30 '20 at 08:27
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$(x+y+z)^2=xyz \implies \Bigl(2 z^2 - y (z^2 - 4 z)\Bigr)^2 - (z^2 - 4 z) \Bigl(2 (x + y + z) - y z\Bigr)^2 = 4 z^4$

Some natural solutions:

(5, 25, 20)
(5, 45, 25)
(5, 100, 45)
(5, 245, 100)
(5, 625, 245)
(5, 1620, 625)
(5, 4225, 1620)
(5, 11045, 4225)
(5, 28900, 11045)
(6, 18, 12)
(6, 48, 18)
(6, 162, 48)
(6, 588, 162)
(6, 2178, 588)
(6, 8112, 2178)
(6, 30258, 8112)
(6, 112908, 30258)
(6, 421362, 112908)
(6, 1572528, 421362)
(6, 5868738, 1572528)
(6, 21902412, 5868738)
(6, 81740898, 21902412)
(6, 305061168, 81740898)
(6, 1138503762, 305061168)
(6, 4248953868, 1138503762)
(6, 15857311698, 4248953868)
(8, 16, 8)
(8, 72, 16)
(8, 400, 72)
(8, 2312, 400)
(8, 13456, 2312)
(8, 78408, 13456)
(8, 456976, 78408)
(8, 2663432, 456976)
(8, 15523600, 2663432)
(9, 36, 9)
(9, 225, 36)
(9, 1521, 225)
(9, 10404, 1521)
(12, 18, 6)
(12, 150, 18)
(12, 1458, 150)
(12, 14406, 1458)
(12, 142578, 14406)
(12, 1411350, 142578)
(12, 13970898, 1411350)
(12, 138297606, 13970898)
(12, 1369005138, 138297606)
(12, 13551753750, 1369005138)
(12, 134148532338, 13551753750)
(12, 1327933569606, 134148532338)
(12, 13145187163698, 1327933569606)
(12, 130123938067350, 13145187163698)
(12, 1288094193509778, 130123938067350)
(12, 12750817997030406, 1288094193509778)
(12, 126220085776794258, 12750817997030406)
(16, 72, 8)
(16, 968, 72)
(16, 13448, 968)
(16, 187272, 13448)
(16, 2608328, 187272)
(16, 36329288, 2608328)
(16, 506001672, 36329288)
(16, 7047694088, 506001672)
Dmitry Ezhov
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    https://math.stackexchange.com/questions/1930438/equation-with-vieta-jumping-xyz2-nxyz – Will Jagy May 29 '20 at 19:38
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    There are four Markov type trees, the other solutions obtained by that Vieta jumping, as in the Markov Numbers tree. The four roots with $x \geq y \geq z \geq 1$ are 1 X: 16 Y: 8 Z: 8 ///// 1 X: 18 Y: 12 Z: 6 ///// 1 X: 25 Y: 20 Z: 5 ///// 1 X: 9 Y: 9 Z: 9 ///// – Will Jagy May 29 '20 at 19:42
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    https://en.wikipedia.org/wiki/Markov_number – Will Jagy May 29 '20 at 19:46
  • From the four cases, $(9,9,9)$ is missing in the above list. – Dietrich Burde May 30 '20 at 08:29