Does there exist a non-trivial rational solution of $2(1+x^2)+4x(y-y^3)^2=z^2$.
This equation might seem very uninteresting to many of you but it has resulted after solving many simultaneous Diophantine systems. My answer depends upon this one.
Edit: I am expecting some parametric family of solutions.
On
The positive $x$ for which both $x$ and $-x$ give a solution come in the sequence $x_0 = 1,$ $x_1 = 7,$ then $$ x_{n+2} = 6 x_{n+1} - x_n. $$ I see, in these cases we have $y-y^3 = 0.$ This suggests trying to gather the other $x$ values by the value of $y$ that works. Hmmm.
Alright, I put $y=3$ and $y = 5$ as separate answers. There are plenty of $y$ that do not work at all. If $y=7,$ we find $y^3 - y = 336.$ This is bad, as $336^4 - 1 = 5 \cdot 17 \cdot 29 \cdot 67 \cdot 229 \cdot 337 $ cannopt be expressed as $w^2 - 2 v^2$ in integers.
jagy@phobeusjunior:~$ ./mse
x: -1 y: -1 z: 2 = 2
x: -1 y: 0 z: 2 = 2
x: -1 y: 1 z: 2 = 2
x: 1 y: -1 z: 2 = 2
x: 1 y: 0 z: 2 = 2
x: 1 y: 1 z: 2 = 2
x: -7 y: -1 z: 10 = 2 5
x: -7 y: 0 z: 10 = 2 5
x: -7 y: 1 z: 10 = 2 5
x: 7 y: -1 z: 10 = 2 5
x: 7 y: 0 z: 10 = 2 5
x: 7 y: 1 z: 10 = 2 5
x: -41 y: -1 z: 58 = 2 29
x: -41 y: 0 z: 58 = 2 29
x: -41 y: 1 z: 58 = 2 29
x: 41 y: -1 z: 58 = 2 29
x: 41 y: 0 z: 58 = 2 29
x: 41 y: 1 z: 58 = 2 29
x: 89 y: 3 z: 470 = 2 5 47
x: 119 y: 3 z: 550 = 2 5^2 11
x: -239 y: -1 z: 338 = 2 13^2
x: -239 y: 0 z: 338 = 2 13^2
x: -239 y: 1 z: 338 = 2 13^2
x: 239 y: -1 z: 338 = 2 13^2
x: 239 y: 0 z: 338 = 2 13^2
x: 239 y: 1 z: 338 = 2 13^2
x: 329 y: 5 z: 4378 = 2 11 199
x: 409 y: 3 z: 1130 = 2 5 113
x: 479 y: 3 z: 1250 = 2 5^4
x: 791 y: 5 z: 6842 = 2 11 311
x: -1241 y: 3 z: 470 = 2 5 47
x: -1271 y: 3 z: 550 = 2 5^2 11
x: -1393 y: -1 z: 1970 = 2 5 197
x: -1393 y: 0 z: 1970 = 2 5 197
x: -1393 y: 1 z: 1970 = 2 5 197
x: 1393 y: -1 z: 1970 = 2 5 197
x: 1393 y: 0 z: 1970 = 2 5 197
x: 1393 y: 1 z: 1970 = 2 5 197
x: -1561 y: 3 z: 1130 = 2 5 113
x: -1631 y: 3 z: 1250 = 2 5^4
x: 2359 y: 3 z: 4070 = 2 5 11 37
x: 2609 y: 3 z: 4430 = 2 5 443
x: -3511 y: 3 z: 4070 = 2 5 11 37
x: 3607 y: 5 z: 15290 = 2 5 11 139
x: -3761 y: 3 z: 4430 = 2 5 443
x: 4639 y: 3 z: 7330 = 2 5 733
x: 4993 y: 5 z: 18370 = 2 5 11 167
x: 5089 y: 3 z: 7970 = 2 5 797
x: -5791 y: 3 z: 7330 = 2 5 733
x: -6241 y: 3 z: 7970 = 2 5 797
x: 7039 y: 5 z: 22462 = 2 11 1021
x: -8119 y: -1 z: 11482 = 2 5741
x: -8119 y: 0 z: 11482 = 2 5741
x: -8119 y: 1 z: 11482 = 2 5741
x: 8119 y: -1 z: 11482 = 2 5741
x: 8119 y: 0 z: 11482 = 2 5741
x: 8119 y: 1 z: 11482 = 2 5741
x: 9041 y: 5 z: 26158 = 2 11 29 41
x: 16369 y: 3 z: 23950 = 2 5^2 479
x: 17489 y: 5 z: 40238 = 2 11 31 59
x: -17521 y: 3 z: 23950 = 2 5^2 479
x: 17839 y: 3 z: 26030 = 2 5 19 137
x: -18991 y: 3 z: 26030 = 2 5 19 137
Alright, that works pretty well. For example, when $y = 3,$ and taking $z = 2t,$ we wind up solving the Pell type $$ (x+576)^2 - 2 t^2 = 577 \cdot 23 \cdot 5^2 $$ There are always details: both $x+576$ and $t$ must be divisible by $5,$ so that $$ x+576 = 5 w, \; t = 5 v, \; w^2 - 2 v^2 = 577 \cdot 23 = 13271 $$ For each $w$ in the output below, we get two $x$ values from $\pm w,$ namely $5w-576$ and $-5w-576.$ The degree two recursion for $x$ is a mess, but we get less trouble with $u,t$ and therefore $z.$ Still intricate.
jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
Automorphism matrix:
3 4
2 3
Automorphism backwards:
3 -4
-2 3
3^2 - 2 2^2 = 1
w^2 - 2 v^2 = 13271 = 23 577
Sun Jan 27 18:25:45 PST 2019
w: 133 v: 47 SEED KEEP +-
w: 139 v: 55 SEED KEEP +-
w: 197 v: 113 SEED BACK ONE STEP 139 , -55
w: 211 v: 125 SEED BACK ONE STEP 133 , -47
w: 587 v: 407
w: 637 v: 443
w: 1043 v: 733
w: 1133 v: 797
w: 3389 v: 2395
w: 3683 v: 2603
w: 6061 v: 4285
w: 6587 v: 4657
w: 19747 v: 13963
w: 21461 v: 15175
w: 35323 v: 24977
w: 38389 v: 27145
w: 115093 v: 81383
w: 125083 v: 88447
w: 205877 v: 145577
w: 223747 v: 158213
w: 670811 v: 474335
w: 729037 v: 515507
w: 1199939 v: 848485
w: 1304093 v: 922133
w: 3909773 v: 2764627
w: 4249139 v: 3004595
w: 6993757 v: 4945333
w: 7600811 v: 5374585
Sun Jan 27 18:26:11 PST 2019
w^2 - 2 v^2 = 13271 = 23 577
jagy@phobeusjunior:~$
jagy@phobeusjunior:~$ ./mse
x: 1883191 y: 40 z: 175564058 = 2 59 101 14731
x: 15226951 y: 40 z: 499630198 = 2 607 411557
x: 28177657 y: 40 z: 680201290 = 2 5 223 305023
x: 35790119 y: 40 z: 766951382 = 2 383475691
x: 61524257 y: 40 z: 1007136670 = 2 5 37 127 21433
x: 72580159 y: 40 z: 1094624642 = 2 41 47 284023
Taking $y=5$ gives $$ (x+ 120^2)^2 - 2 t^2 = 120^4 - 1 = 11^2 \cdot 7 \cdot 17 \cdot 14401 $$ so $x + 120^2 = \pm 11 w$ and $t = 11 v$ and $$ w^2 - 2 v^2 = 7 \cdot 17 \cdot 14401 = 1713719$$