Pell number factorization and divisibility question In a problem I’m working on, I have positive integers $a,b,c,d$ satisfying
$$
(ab)^2-2(cd)^2=1.  \tag{1}
$$
(So evidently $cd$ is a Pell number, and $ab$ is its companion.) Furthermore, say the following divisibility conditions also hold:
\begin{align}
a &\mid (b^2+2c^2) \\
b &\mid (a^2-2d^2) \\ \tag{2}
c &\mid (a^2+d^2) \\
d &\mid (b^2-c^2).
\end{align}
A brute-force computer search turns up only two solutions so far: $(a,b,c,d)=(3,1,1,2)$ and $(a,b,c,d)=(3,1,2,1)$. I believe they may be the only ones.
I’ve tried to algebraically prove that conjecture, to no avail. From the divisibility system, we see that $a$ must be of the form $x^2+2y^2$, and $c$ must be of the form $u^2+v^2$. From $(1),$ we deduce, for example, that $\gcd(ab,cd)=1$, and also $ab=r^2+2s^2$ (so $b$ is also of that form) and $cd=2rs$ with $r^2-2s^2=\pm1$. But I can’t seem to find the magic combination to complete the proof.
Any hints, assistance, or references would be appreciated.
 A: EDIT
Incomplete solution.
From $(1)$ it is clear that $ab$ is odd.
$$\begin{aligned}(ab)^2-2(cd)^2=1 \quad &\Longleftrightarrow \quad (ab)^2-1=2(cd)^2\\&\Longleftrightarrow \quad (\underbrace{ab-1}_{2k})(\underbrace{ab+1}_{2k+2})=2(cd)^2, \end{aligned} \tag{*}$$
it follows that $cd$ is even.

*

*Assume that $cd$ is a power of $2.$ Then $(*)$ rewrites
$$4k(k+1)=2^{2n+1},$$ which is only possible if $k=1.$ Taking into account $(2),$ this gives the two solutions you already found.


*In general, $(*)$ simplifies to
$$2k(k+1)=(cd)^2.$$
Let $p\neq 2$ be a prime. Between two consecutive numbers, at most one is divisible by $p.$ However, one of them can be divisible by $p^{2},$ as pointed out by Arthur Vause.

*

*As an example, $k=8$ satisfies and gives  $$(a,b,c,d), \; \text{where}\; ab=17 \;\text{ and} \;cd=12.$$ However, $(2)$ is not checked.

*A different family of solutions could be obtained when $k=49.$ Here $$(a,b,c,d) \; \text{are such that}\; ab=99 \;\text{ and} \;cd=70.$$
A: Prompted by a suggestion from @gnasher729 I have analysed the solutions to (1) that satisfy one or more of (2) where the larger of a,b or the larger of c,d divides the relevant expression of (2).
Defining $(x_k, y_k)$ to be the k-th solution to $x^2-2y^2=1$,
so for $k=0$ to $5$, $(x_k, y_k) =  (1,0), (3, 2),(17, 12),(99, 70),(577, 408),(3363, 2378) $
The vast majority of results are where the smaller of a,b = 1 or the smaller of c,d = 1.
Apart for a small number of results for $(x_3,y_3),(x_6,y_6),(x_9,y_9)$, the only other results occur for odd values of k when $x_k$ is of the form $m^2−1$, and $(a,b)=(m+1,m−1)$ or $(m−1,m+1)$.
Results are available here: https://drive.google.com/file/d/1ldFWhK1Dm83roP-1oRYT59RLdUNt-Ps9/view
I terminated the search at $k=75$, not because the factorisation of $x_k$ and $y_k$ is difficult, bu simply because the number of divisors was very large, giving over 2 billion combinations of $a,b,c,d$
The "erratic" solutions for $k=3,6,9$ are:
k = 3, (x,y) = (99, 70):
large d divides b^2 - c^2, (a,b,c,d)=  (3, 33, 2, 35)
large d divides b^2 - c^2, (a,b,c,d)=  (3, 33, 5, 14)
large d divides b^2 - c^2, (a,b,c,d)=  (3, 33, 7, 10)
large d divides b^2 - c^2, (a,b,c,d)=  (33, 3, 7, 10)
k = 6, (x,y) = (19601, 13860):
large d divides b^2 - c^2, (a,b,c,d)=  (1153, 17, 28, 495)
large d divides b^2 - c^2, (a,b,c,d)=  (17, 1153, 35, 396)
k = 9, (x,y) = (3880899, 2744210):
large d divides b^2 - c^2, (a,b,c,d)=  (179, 21681, 985, 2786)
A: The problem $x^2 - 2y^2 = + / - 1$ is well known: the first solutions are (1, 0), (1, 1), (3, 2), (7, 5), (17, 12) etc. with (x, y) followed by (x + 2y, x + y). The recurrence can be reversed, so you get from one solution to the previous one, which shows that there are no other solutions.
With this sequence the remainder alternates between 1 and -1, so we take every second term (1, 0), (3, 2), (17, 12), (99, 70) etc. with (x, y) followed by (3x + 4y, 2x+3y).
Factoring x, y gives you a, b, c, d so you can check your other conditions. x grows slightly more than a factor 5.8 from one solution to the next (factor is about 3 + sqrt 8), so you should find all solutions that you can factor in reasonable time quite quickly; there are about 50 pairs (x, y) less than $2^{128}$.
The chance that one of a, b, c, d divide a random integer is 1/a, 1/b, 1/c, 1/d. The chance of all four dividing four random integers is 1/xy. So this is very unlikely to happen unless you find a reason why they should divide. The chance for (17, 12) is only one in 204 and things get worse quickly.
