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.

  • 2
    $\begingroup$ It may be worth noting that $4r^4+1=(2r^2+2r+1)(2r^2-2r+1)$. $\endgroup$ – Gerry Myerson Oct 18 '13 at 11:30
  • 1
    $\begingroup$ @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 Oct 18 '13 at 11:43
  • 1
    $\begingroup$ 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 Oct 18 '13 at 11:48
  • 2
    $\begingroup$ @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 Oct 18 '13 at 14:10
  • $\begingroup$ @Lucia: Excellent point — thanks for the correction! $\endgroup$ – Kieren MacMillan Oct 18 '13 at 14:15

[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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.