I'm rereading some older unfinished material again. I had the following problem which I still can only access by brute force and would like to -at least- understand more about an analytical access.
Consider $(r,s,t) \in \mathbb N^+ $ .
I ask whether for the following set of modular equations there is any option to find a general key to describe the set of solutions.
Consider
$$ \begin{array} {}
r&+s&+rs & \equiv 0 &\pmod t \\
s&+t&+st & \equiv 0 &\pmod r \\
t&+r&+tr & \equiv 0 &\pmod s \\
\end{array} \tag 1$$
To have this algebraically better(?) accessible I introduce $i,j,k \in \mathbb N^+$ and rewrite
$$ \begin{array} {}
r&+s&+rs & = i \cdot t \\
s&+t&+st & = j \cdot r \\
t&+r&+tr & = k \cdot s \\
\end{array} \tag 2$$
I hoped it would be possible to define some matrix-operation, but with modular arithmetic, which would help solve this (or at least reformulate in a more intuitive scheme), but I ran into too many problems so far - so this is one aspect of my question:
Can the diophantine problem (1) be handled by matrix-operations, helpful for the finding of a general solution? (For instance: how to implement "modular matrix-inversion" if possible at all...)
I think I found correctly, that all of $(r,s,t)$ must be even. Beside this I did not really proceed algebraically and only have a set of solutions (which seems to be infinite) by enumerating solutions. If I begin with setting $r$ and $s$ then the set of solutions for the third unknown $t$ is finite or even empty.
Can the set of solutions be given as parametrical description? (We have perhaps something similar to the sets of solutions of the Pell-problem)
(Remark on background: if $(r+1,s+1,t+1)$ are all primes then $n=(r+1)(s+1)(t+1)$ should be a Carmichael number, btw. - this is also the origin for my fiddling)
Beginning of set of solutions:
r s t fixing r,s, finding t. s is always been checked up to 2^15
==================
2 2 2
2 2 4
2 2 8
2 4 14
2 8 26
2 10 16 likely complete for r=2
------------------
4 4 4
4 4 8
4 4 12
4 4 24
4 6 34
4 8 44
4 12 16
4 12 64
4 16 28
4 24 124
4 28 72 likely complete for r=4
------------------
6 6 6
6 6 12
6 6 24
6 6 48
6 12 18
6 12 30
6 12 90
6 16 118
6 18 66
6 22 40
6 24 174
6 30 72
6 48 342
6 54 192
6 72 102 likely complete for r=6
------------------
8 8 8
8 8 16
8 8 40
8 8 80
8 10 98
8 16 152
8 20 188
8 28 52
8 40 368
8 80 728
8 88 400 likely complete for r=8
------------------
10 10 10
10 10 20
10 10 30
10 10 40
10 10 60
10 10 120
10 20 230
10 30 340
10 40 50
10 40 90
10 40 450
10 50 140
10 60 670
10 70 260
10 120 1330
10 130 720 likely complete for r=10
------------------
12 12 12
12 12 24
12 12 84
12 12 168
12 24 36
12 24 108
12 24 324
12 28 376
12 36 60
12 36 96
12 36 240
12 40 76
12 42 558
12 60 396
12 84 1104
12 96 420
12 132 192
12 168 2196
12 180 1176
12 216 564
12 276 360 likely complete for r=12
------------------
14 14 14
14 14 28
14 14 56
14 14 112
14 14 224
14 28 434
14 56 854
14 112 1694
14 224 3374
14 238 1792 likely complete for r=14
------------------
16 16 16
16 16 32
16 16 48
16 16 96
16 16 144
16 16 288
16 24 424
16 32 80
16 32 112
16 32 560
16 40 232
16 48 64
16 48 208
16 48 832
16 56 88
16 64 368
16 72 1240
16 96 1648
16 112 480
16 144 2464
16 160 912
16 288 4912
16 304 2592
16 352 1200
16 432 736 likely complete for r=16
------------------
18 18 18
18 18 36
18 18 72
18 18 90
18 18 180
18 18 360
18 36 54
18 36 234
18 36 702
18 42 408
18 48 186
18 54 522
18 72 1386
18 90 108
18 90 288
18 90 1728
18 108 414
18 126 1206
18 180 3438
18 198 270
18 198 1260
18 270 468
18 360 6858
18 378 3600 likely complete for r=18
------------------
Above simultaneous equation shown below:
$\begin{array} {} r&+s&+rs & = i \cdot t \\ s&+t&+st & = j \cdot r \\ t&+r&+tr & = k \cdot s \\ \end{array} \tag 1$
Equation, $(1)$ has parametric solution & is shown below:
$(i,j,k)=[(4m+1),(36m+5),(9m+2)]$
$(r,s,t)=(6m,12m,18m)$
for, $m=4$ we get:
$(r,s,t)=(24,48,72)$ &
$(i,j,k)=(17,149,38)$