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In a table of fundamental solutions $f_1(x,y)$ to Pell equations,

$$x^2-dy^2=1\tag1$$

with $d<110$, two will stand out,

$$(U_{61})^6 = \big(\tfrac{39+5\sqrt{61}}{2}\big)^6 = x+y\sqrt{61} =1766319049+226153980\sqrt{61}$$

$$(U_{109})^6 = \big(\tfrac{261+25\sqrt{109}}{2}\big)^6 = x+y\sqrt{109} = 158070671986249+15140424455100\sqrt{109}$$

where $U_d$ is a fundamental unit. It is a theorem that if fundamental solution {$u,v$} to,

$$u^2-dv^2=-4\tag2$$

is odd, then $f_1(x,y)$ for $(1)$ is given by $\big(\tfrac{u+v\sqrt{d}}{2}\big)^6 = x+y\sqrt{d}\,$ and, being sixth powers, are generally big. A necessary (but not sufficient) condition is that $d=8m+5$.

In "Large fundamental solutions of Pell's equation", the authors point out that,

$$d=60n'+1$$

$$d=60n'+49$$

have record sizes. Together with the condition $d=8m+5$, then $n'$ must be odd, so,

$$d = 120n+\color{brown}{61}$$

$$d = 120n+\color{brown}{109}$$

hence those two stand-outs are just the first in a rather orderly sequence.

Q: What is the necessary condition such that prime $d = 120n+61$ (or $120n+109$) has odd fundamental solution $u,v$ to $u^2-dv^2=-4$?

I checked that, for primes, the list of $d=120n+61$ with odd solutions {$u,v$} is,

$$d_1 = 61, 181, 421, 541, 661, 1021, 1381, 1621, 1741, 2341, 3061,\dots$$

while those with even {$u,v$},

$$d_2 = 1861, 2221, 3181, 3301, 4021,\dots$$

and it seems that, within a bound, majority belong to the former. So what else (mod constraints?) distinguishes $d_1$ from $d_2$?

Addendum:

For those interested, sequence A107997 is square-free $d$ such that $u,v$ are both odd. Excepting $p = 5$, the rest modulo $120$ fall into eight classes. The best $d<10000$ per class are given below:

$$\begin{array}{|c|c|c|c|c|} \text{#}&\text{Form}&d&\text{digits of}\; u&\tfrac{1}{6}\sqrt{d}\,\ln(\ln(d))\\ 1&120n+13&9013&23&-\\ 2&120n+29&9629&18&-\\ 3&120n+37&8317&23&-\\ 4&120n+53&8933&13&-\\ 5&\color{brown}{120n+61}&8941&\color{brown}{34}&\color{brown}{34.8}\\ 6&120n+77&11117&11&-\\ 7&120n+101&9461&20&-\\ 8&\color{brown}{120n+109}&9949&\color{brown}{36}&\color{brown}{36.9}\\ \end{array}$$

The winners, hands down, are $d=120n+61,\;120n+109$ and they alone approach the website authors' asymptotics. (The sixth class was so poor I extended it to $d<12000$.)

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    $\begingroup$ This question is very interesting, I think. But what is the reason for restricting to prime $d$? $\endgroup$ – Travis Jun 24 '15 at 7:55
  • $\begingroup$ @Travis: Good point. The authors of the linked website limited themselves to prime $d$ so I just followed their lead. $\endgroup$ – Tito Piezas III Jun 24 '15 at 7:58
  • $\begingroup$ Ah, I see. The authors write, "We found empirically that for $D>394$ all record-holders appear to be congruent, modulo $60$, to either $1$ or $49$. Furthermore, for these values of $D$ all record-holders appear to be prime numbers." It's certainly not clear to me whether record-holder status should have some bearing on the parity issue. The authors' power-law observation (especially the conjecture that the exponent is $\log 2 / \log \phi$) is quite interesting, though I suspect this might have a reasonably elementary explanation. $\endgroup$ – Travis Jun 24 '15 at 8:06
  • $\begingroup$ Champions is always a Prime number. Prime numbers are associated with solutions of Pell equations. $\endgroup$ – individ Jun 24 '15 at 8:14
  • $\begingroup$ Small numbers that are $\equiv 1,49 \pmod {60}$ are quite likely to be prime as you already can't have factors $2,3,5$ $\endgroup$ – Ross Millikan Jun 24 '15 at 14:26

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