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.

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