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Today I proved finiteness for the problem here: Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?

namely: IF we have integers $ \color{red}{x,y,t \geq 1,}$ with $ \color{red}{xy > t}$ and $ \color{red}{q \geq 3}$ such that $$ q = \frac{x^2 + y^2}{xy-t}, $$ THEN $$ q \leq 1 + \frac{324}{25} t^2. $$ That is, for a given value of $t,$ there are finitely many values of $q$ that work.

Given the bounds I found, it is extremely quick to check for solutions by computer. I have done so, and have a simple conjecture, massively komputer konfirmed: $$ \color{red}{ q \leq t^2 + 2 t + 2}. $$

QUESTION: can anyone prove the conjecture? It is true for $t \leq 8192.$

Oh: there is always such a solution for any $t,$ with $x = t+1, y=1.$

The main contribution of the way Hurwitz wrote this sort of thing, pretty much the same idea as Vieta Jumping, is that, if there is any solution with some $t,q,$ then there is a solution with $2x \leq qy$ and $2y \leq qx.$ I have a pdf of his article. The strongest item i got, for such a Hurwitz Grundlösung, was $xy \leq 4t.$ That is mostly from Lagrange multipliers, i think the conjecture needs better, more Diophantine methods.

Here are some samples, because of symmetry I printed only $x \geq y:$

 x     2     y     1    t      1    q      5  +++  

 x     2     y     2    t      2    q      4
 x     3     y     1    t      2    q     10  +++  

 x     2     y     2    t      3    q      8
 x     3     y     3    t      3    q      3
 x     4     y     1    t      3    q     17  +++  
 x     4     y     2    t      3    q      4
 x     5     y     1    t      3    q     13

 x     4     y     2    t      4    q      5
 x     5     y     1    t      4    q     26  +++  

 x     3     y     2    t      5    q     13
 x     6     y     1    t      5    q     37  +++  
 x     7     y     1    t      5    q     25

 x     3     y     3    t      6    q      6
 x     4     y     2    t      6    q     10
 x     7     y     1    t      6    q     50  +++  

Why not? here are the 62 fundamental solutions with $x \geq y$ for $t=8192:$

 x    96     y    96    t   8192    q     18
 x   100     y    84    t   8192    q     82
 x   108     y    76    t   8192    q   1090
 x   128     y   128    t   8192    q      4
 x   152     y    56    t   8192    q     82
 x   158     y    54    t   8192    q     82
 x   176     y    48    t   8192    q    130
 x   188     y    44    t   8192    q    466
 x   192     y    64    t   8192    q     10
 x   228     y    36    t   8192    q   3330
 x   240     y    48    t   8192    q     18
 x   241     y    37    t   8192    q     82
 x   267     y    31    t   8192    q    850
 x   277     y    33    t   8192    q     82
 x   288     y    32    t   8192    q     82
 x   344     y    24    t   8192    q   1858
 x   397     y    21    t   8192    q   1090
 x   416     y    32    t   8192    q     34
 x   420     y    20    t   8192    q    850
 x   440     y    24    t   8192    q     82
 x   526     y    22    t   8192    q     82
 x   528     y    16    t   8192    q   1090
 x   551     y    15    t   8192    q   4162
 x   590     y    14    t   8192    q   5122
 x   592     y    16    t   8192    q    274
 x   633     y    13    t   8192    q  10834
 x   661     y    13    t   8192    q   1090
 x   684     y    12    t   8192    q  29250
 x   716     y    12    t   8192    q   1282
 x   796     y    12    t   8192    q    466
 x   804     y    20    t   8192    q     82
 x   912     y    16    t   8192    q    130
 x   913     y     9    t   8192    q  33346
 x  1032     y     8    t   8192    q  16642
 x  1064     y     8    t   8192    q   3538
 x  1171     y     7    t   8192    q  274258
 x  1251     y     7    t   8192    q   2770
 x  1256     y     8    t   8192    q    850
 x  1366     y     6    t   8192    q  466498
 x  1374     y     6    t   8192    q  36306
 x  1614     y     6    t   8192    q   1746
 x  1928     y     8    t   8192    q    514
 x  1942     y     6    t   8192    q   1090
 x  2052     y     4    t   8192    q  263170
 x  2068     y     4    t   8192    q  53458
 x  2100     y     4    t   8192    q  21202
 x  2196     y     4    t   8192    q   8146
 x  2308     y     4    t   8192    q   5122
 x  2484     y     4    t   8192    q   3538
 x  2759     y     3    t   8192    q  89554
 x  2788     y     4    t   8192    q   2626
 x  2899     y     3    t   8192    q  16642
 x  3303     y     3    t   8192    q   6354
 x  3972     y     4    t   8192    q   2050
 x  4098     y     2    t   8192    q  4198402
 x  4890     y     2    t   8192    q  15058
 x  8066     y     2    t   8192    q   8194
 x  8193     y     1    t   8192    q  67125250  +++  
 x  8245     y     1    t   8192    q  1282642
 x  8977     y     1    t   8192    q  102658
 x  9805     y     1    t   8192    q  59602
 x 16257     y     1    t   8192    q  32770
Will Jagy
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  • Actually if $q<0$ the solutions are endless. This task is solved easily, as the binary quadratic form reduces to the Pell equation. But you still stubbornly use too much. I can not understand why make the computer all the numbers to sort? – individ Oct 10 '15 at 17:56
  • @individ, quoting from above: IF we have integers $x,y,t \geq 1,$ with $xy > t$ and $q \geq 3$ such that $$ q = \frac{x^2 + y^2}{xy-t}, $$ – Will Jagy Oct 10 '15 at 18:19
  • @individ, the original question in June 2014 was not careful, http://math.stackexchange.com/questions/829228/is-it-true-that-fx-y-dfracx2y2xy-t-has-only-finitely-many-distinct-i She did not say that she was only interested in positive ratios $(x^2 + y^2)/(xy-t).$ However, she did say that an upper bound on the ratio would give finiteness. This means that she was, in fact, thinking only about positive ratios. In particular, she did not include $x=1,y=1,t=2.$ In any case, I am quite definitely considering *positive ratios only*. – Will Jagy Oct 10 '15 at 19:11
  • I there is presented a General formula which reduces to the Pell equation. And necessary condition for the existence of solutions is represented by the root must be rational. I can't understand why these formulas You don't like? – individ Oct 11 '15 at 04:15
  • It seems you have answered this question yourself here. Should this be closed as a duplicate? – Servaes Jun 12 '19 at 18:40

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