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While working on another problem (Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$), I came across a question which seems to be of [semi-] independent interest.

Conjecture. If $r > s \ge 1$ are integers such that $r$ is odd and $s$ is even and $$(r+s)^2 \mid (4r^4+1), \qquad(\star)$$ then $(r,s) \in \{(3,2), (21,8), (119,50), (697,288), (2679,910), (4059,1682), \dots\}$.

Note that these are essentially Pell oblongs (i.e., $r$ is http://oeis.org/A001652) and doubles of Pell squares (i.e., $s$ is http://oeis.org/A114619), except for the outlier solution $(2679,910)$, which is the only one maxima gave for $r \le 11000$ (I'm increasing that bound now).

I haven't the foggiest notion how to prove any conditions on $r$ and $s$ satisfying $(\star)$, or to figure out why there are any 'outlier' solutions at all (and, of course, if there are any more). EDIT: For example, it seems that all solutions (including the 'outlier') with $r>3$ satisfy $s/r < 1/2$; can one prove this bound?

Any suggestions, hints, full solutions, etc. greatly appreciated.

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    $\begingroup$ It may be worth noting that $4r^4+1=(2r^2+2r+1)(2r^2-2r+1)$. $\endgroup$
    – Gerry Myerson
    Commented Oct 18, 2013 at 11:30
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    $\begingroup$ @GerryMyerson: Thanks! I recalled that just after I posted… Since those two factors are odd, they are evidently relatively prime. Hence any square dividing $4r^4+1$ divides exactly one of $2r^2+1 \pm 2r$ (i.e., the square is not "split" across the two factors). Now if $s/r > 1/2$, then $(r+s)^2 > (3r/2)^2 = (9/4)r^2$, and hence this [alleged] square factor $2r^2 < (r+s)^2 \mid (2r^2+1\pm 2r)$. Unless $(r+s)^2 = (2r^2+1+2r)$, wouldn't that be an immediate contradiction? In other words, is that a valid proof of the condition $s/r < 1/2$ for $r > 3$? $\endgroup$
    – Kieren MacMillan
    Commented Oct 18, 2013 at 11:43
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    $\begingroup$ Looks that way. Also, letting $r+s=y$, we have $y^2=2r^2\pm2r+1$, which leads to a couple of Pellians, which will give you many (but apparently not all) solutions. $\endgroup$
    – Gerry Myerson
    Commented Oct 18, 2013 at 11:48
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    $\begingroup$ @KierenMacMillan: I don't see why you can't have $(r+s)=ab$ and then $a^2|(2r^2+1+2r)$ and $b^2|(2r^2+1-2r)$. For any prime square your statement is fine. $\endgroup$
    – Lucia
    Commented Oct 18, 2013 at 14:10
  • $\begingroup$ @Lucia: Excellent point — thanks for the correction! $\endgroup$
    – Kieren MacMillan
    Commented Oct 18, 2013 at 14:15

1 Answer 1

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[corrected $-$ see edit history for previous attempt]

The conjecture is false: there are infinitely many "Pell" parametrizations, some with larger values of $s/r$. For example, $$ (r,s) = (307470495089672071303, \, 295528756570432706202) $$ has $s/r \sim .961$.

This was obtained as follows. Recall that $4r^4 + 1$ factors as $(2r^2-2r+1) (2r^2+2r+1)$. Start from the first solution $(r,s) = (3,2)$, with $2r^2-2r+1 = 13$ and $2r^2+2r+1 = 5^2$. Instead of generalizing to $2r^2+2r+1 = y^2$, we generalize to $2r^2-2r+1 = 13y^2$ and $2r^2+2r+1 \equiv 0 \bmod 25$. This is a Fermat-Pell equation with a congruence condition, and since we have one solution $(r,y) = (3,2)$ there must be infinitely many others. The equation is $x^2-26y^2=1$ with $x=2r-1$, which must be positive and $1 \bmod 4$ to satisfy the sign and parity conditions on $n$. The general solution is $x + \sqrt{26} \, y = (5 + \sqrt{26})^{4k+1}$ ($k=0,1,2,\ldots$), and then the ${}\bmod 25$ condition gives $5|k$. The solution displayed above comes from $k=5$.

We can obtain further infinite families by iterating trick of switching between the $2r^2-2r+1$ and $2r^2+2r+1$ factors, and by starting from some other solution such as the $r=2679$ "outlier" or any other solution that a numerical search might find.

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