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Rational solutions of $2(1+x^2)+4x(y-y^3)^2=z^2$
0
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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.
number-theory diophantine-equations
asked Jan 28 at 0:11
ershersh
438113
$endgroup$
add a comment |
0
$begingroup$
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.
number-theory diophantine-equations
asked Jan 28 at 0:11
ershersh
438113
$endgroup$
2
$begingroup$
$(x,y,z)=(7,1,10)$ works (by a simple search).
$endgroup$
– lulu
Jan 28 at 0:20
$begingroup$
Good! Though I am expecting some parametric solution.
$endgroup$
– ersh
Jan 28 at 0:23
$begingroup$
Then you should edit your post to ask for that.
$endgroup$
– lulu
Jan 28 at 0:23
$begingroup$
Thanks!. Done!.
$endgroup$
– ersh
Jan 28 at 0:25
add a comment |
0
0
0
1
$begingroup$
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.
number-theory diophantine-equations
asked Jan 28 at 0:11
ershersh
438113
$endgroup$
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.
number-theory diophantine-equations
number-theory diophantine-equations
asked Jan 28 at 0:11
ershersh
438113
asked Jan 28 at 0:11
ershersh
438113
edited Jan 28 at 0:24
ersh
asked Jan 28 at 0:11
ershersh
438113
asked Jan 28 at 0:11
ershersh
438113
asked Jan 28 at 0:11
ershersh
438113
438113
2
$begingroup$
$(x,y,z)=(7,1,10)$ works (by a simple search).
$endgroup$
– lulu
Jan 28 at 0:20
$begingroup$
Good! Though I am expecting some parametric solution.
$endgroup$
– ersh
Jan 28 at 0:23
$begingroup$
Then you should edit your post to ask for that.
$endgroup$
– lulu
Jan 28 at 0:23
$begingroup$
Thanks!. Done!.
$endgroup$
– ersh
Jan 28 at 0:25
add a comment |
2
$begingroup$
$(x,y,z)=(7,1,10)$ works (by a simple search).
$endgroup$
– lulu
Jan 28 at 0:20
$begingroup$
Good! Though I am expecting some parametric solution.
$endgroup$
– ersh
Jan 28 at 0:23
$begingroup$
Then you should edit your post to ask for that.
$endgroup$
– lulu
Jan 28 at 0:23
$begingroup$
Thanks!. Done!.
$endgroup$
– ersh
Jan 28 at 0:25
2
2
$begingroup$
$(x,y,z)=(7,1,10)$ works (by a simple search).
$endgroup$
– lulu
Jan 28 at 0:20
$begingroup$
$(x,y,z)=(7,1,10)$ works (by a simple search).
$endgroup$
– lulu
Jan 28 at 0:20
$begingroup$
Good! Though I am expecting some parametric solution.
$endgroup$
– ersh
Jan 28 at 0:23
$begingroup$
Good! Though I am expecting some parametric solution.
$endgroup$
– ersh
Jan 28 at 0:23
$begingroup$
Then you should edit your post to ask for that.
$endgroup$
– lulu
Jan 28 at 0:23
$begingroup$
Then you should edit your post to ask for that.
$endgroup$
– lulu
Jan 28 at 0:23
$begingroup$
Thanks!. Done!.
$endgroup$
– ersh
Jan 28 at 0:25
$begingroup$
Thanks!. Done!.
$endgroup$
– ersh
Jan 28 at 0:25
add a comment |
4 Answers
4
active
oldest
votes
1
$begingroup$
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
$$ w^2 - 2 v^2 = 7 cdot 17 cdot 14401 = 1713719$$
jagy@phobeusjunior:~$ date
Sun Jan 27 18:51:05 PST 2019
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 = 1713719 = 7 17 14401
Sun Jan 27 18:51:08 PST 2019
w: 1339 v: 199 SEED KEEP +-
w: 1381 v: 311 SEED KEEP +-
w: 1637 v: 695 SEED KEEP +-
w: 1763 v: 835 SEED KEEP +-
w: 1949 v: 1021 SEED BACK ONE STEP 1763 , -835
w: 2131 v: 1189 SEED BACK ONE STEP 1637 , -695
w: 2899 v: 1829 SEED BACK ONE STEP 1381 , -311
w: 3221 v: 2081 SEED BACK ONE STEP 1339 , -199
w: 4813 v: 3275
w: 5387 v: 3695
w: 7691 v: 5359
w: 8629 v: 6031
w: 9931 v: 6961
w: 11149 v: 7829
w: 16013 v: 11285
w: 17987 v: 12685
w: 27539 v: 19451
w: 30941 v: 21859
w: 44509 v: 31459
w: 50011 v: 35351
w: 57637 v: 40745
w: 64763 v: 45785
w: 93179 v: 65881
w: 104701 v: 74029
w: 160421 v: 113431
w: 180259 v: 127459
w: 259363 v: 183395
w: 291437 v: 206075
w: 335891 v: 237509
w: 377429 v: 266881
w: 543061 v: 384001
w: 610219 v: 431489
w: 934987 v: 661135
w: 1050613 v: 742895
w: 1511669 v: 1068911
w: 1698611 v: 1201099
w: 1957709 v: 1384309
w: 2199811 v: 1555501
w: 3165187 v: 2238125
w: 3556613 v: 2514905
w: 5449501 v: 3853379
w: 6123419 v: 4329911
w: 8810651 v: 6230071
w: 9900229 v: 7000519
w: 11410363 v: 8068345
w: 12821437 v: 9066125
Sun Jan 27 18:51:33 PST 2019
w^2 - 2 v^2 = 1713719 = 7 17 14401
jagy@phobeusjunior:~$
answered Jan 28 at 2:53
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
I will work with what you have provided. I will be back.
$endgroup$
– ersh
Jan 28 at 2:58
1
$begingroup$
@ersh after $5,$ the next few $y$ that work are $11,15,21.$ Also, $y=40$ works, so some even $y$ are possible.
$endgroup$
– Will Jagy
Jan 28 at 3:21
add a comment |
2
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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
answered Jan 28 at 1:34
Will JagyWill Jagy
104k5102201
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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:~$
answered Jan 28 at 2:24
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
1
$begingroup$
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
answered Jan 28 at 3:36
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
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
$$ w^2 - 2 v^2 = 7 cdot 17 cdot 14401 = 1713719$$
jagy@phobeusjunior:~$ date
Sun Jan 27 18:51:05 PST 2019
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 = 1713719 = 7 17 14401
Sun Jan 27 18:51:08 PST 2019
w: 1339 v: 199 SEED KEEP +-
w: 1381 v: 311 SEED KEEP +-
w: 1637 v: 695 SEED KEEP +-
w: 1763 v: 835 SEED KEEP +-
w: 1949 v: 1021 SEED BACK ONE STEP 1763 , -835
w: 2131 v: 1189 SEED BACK ONE STEP 1637 , -695
w: 2899 v: 1829 SEED BACK ONE STEP 1381 , -311
w: 3221 v: 2081 SEED BACK ONE STEP 1339 , -199
w: 4813 v: 3275
w: 5387 v: 3695
w: 7691 v: 5359
w: 8629 v: 6031
w: 9931 v: 6961
w: 11149 v: 7829
w: 16013 v: 11285
w: 17987 v: 12685
w: 27539 v: 19451
w: 30941 v: 21859
w: 44509 v: 31459
w: 50011 v: 35351
w: 57637 v: 40745
w: 64763 v: 45785
w: 93179 v: 65881
w: 104701 v: 74029
w: 160421 v: 113431
w: 180259 v: 127459
w: 259363 v: 183395
w: 291437 v: 206075
w: 335891 v: 237509
w: 377429 v: 266881
w: 543061 v: 384001
w: 610219 v: 431489
w: 934987 v: 661135
w: 1050613 v: 742895
w: 1511669 v: 1068911
w: 1698611 v: 1201099
w: 1957709 v: 1384309
w: 2199811 v: 1555501
w: 3165187 v: 2238125
w: 3556613 v: 2514905
w: 5449501 v: 3853379
w: 6123419 v: 4329911
w: 8810651 v: 6230071
w: 9900229 v: 7000519
w: 11410363 v: 8068345
w: 12821437 v: 9066125
Sun Jan 27 18:51:33 PST 2019
w^2 - 2 v^2 = 1713719 = 7 17 14401
jagy@phobeusjunior:~$
answered Jan 28 at 2:53
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
I will work with what you have provided. I will be back.
$endgroup$
– ersh
Jan 28 at 2:58
1
$begingroup$
@ersh after $5,$ the next few $y$ that work are $11,15,21.$ Also, $y=40$ works, so some even $y$ are possible.
$endgroup$
– Will Jagy
Jan 28 at 3:21
add a comment |
1
$begingroup$
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
$$ w^2 - 2 v^2 = 7 cdot 17 cdot 14401 = 1713719$$
jagy@phobeusjunior:~$ date
Sun Jan 27 18:51:05 PST 2019
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 = 1713719 = 7 17 14401
Sun Jan 27 18:51:08 PST 2019
w: 1339 v: 199 SEED KEEP +-
w: 1381 v: 311 SEED KEEP +-
w: 1637 v: 695 SEED KEEP +-
w: 1763 v: 835 SEED KEEP +-
w: 1949 v: 1021 SEED BACK ONE STEP 1763 , -835
w: 2131 v: 1189 SEED BACK ONE STEP 1637 , -695
w: 2899 v: 1829 SEED BACK ONE STEP 1381 , -311
w: 3221 v: 2081 SEED BACK ONE STEP 1339 , -199
w: 4813 v: 3275
w: 5387 v: 3695
w: 7691 v: 5359
w: 8629 v: 6031
w: 9931 v: 6961
w: 11149 v: 7829
w: 16013 v: 11285
w: 17987 v: 12685
w: 27539 v: 19451
w: 30941 v: 21859
w: 44509 v: 31459
w: 50011 v: 35351
w: 57637 v: 40745
w: 64763 v: 45785
w: 93179 v: 65881
w: 104701 v: 74029
w: 160421 v: 113431
w: 180259 v: 127459
w: 259363 v: 183395
w: 291437 v: 206075
w: 335891 v: 237509
w: 377429 v: 266881
w: 543061 v: 384001
w: 610219 v: 431489
w: 934987 v: 661135
w: 1050613 v: 742895
w: 1511669 v: 1068911
w: 1698611 v: 1201099
w: 1957709 v: 1384309
w: 2199811 v: 1555501
w: 3165187 v: 2238125
w: 3556613 v: 2514905
w: 5449501 v: 3853379
w: 6123419 v: 4329911
w: 8810651 v: 6230071
w: 9900229 v: 7000519
w: 11410363 v: 8068345
w: 12821437 v: 9066125
Sun Jan 27 18:51:33 PST 2019
w^2 - 2 v^2 = 1713719 = 7 17 14401
jagy@phobeusjunior:~$
answered Jan 28 at 2:53
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
I will work with what you have provided. I will be back.
$endgroup$
– ersh
Jan 28 at 2:58
1
$begingroup$
@ersh after $5,$ the next few $y$ that work are $11,15,21.$ Also, $y=40$ works, so some even $y$ are possible.
$endgroup$
– Will Jagy
Jan 28 at 3:21
add a comment |
1
1
1
$begingroup$
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
$$ w^2 - 2 v^2 = 7 cdot 17 cdot 14401 = 1713719$$
jagy@phobeusjunior:~$ date
Sun Jan 27 18:51:05 PST 2019
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 = 1713719 = 7 17 14401
Sun Jan 27 18:51:08 PST 2019
w: 1339 v: 199 SEED KEEP +-
w: 1381 v: 311 SEED KEEP +-
w: 1637 v: 695 SEED KEEP +-
w: 1763 v: 835 SEED KEEP +-
w: 1949 v: 1021 SEED BACK ONE STEP 1763 , -835
w: 2131 v: 1189 SEED BACK ONE STEP 1637 , -695
w: 2899 v: 1829 SEED BACK ONE STEP 1381 , -311
w: 3221 v: 2081 SEED BACK ONE STEP 1339 , -199
w: 4813 v: 3275
w: 5387 v: 3695
w: 7691 v: 5359
w: 8629 v: 6031
w: 9931 v: 6961
w: 11149 v: 7829
w: 16013 v: 11285
w: 17987 v: 12685
w: 27539 v: 19451
w: 30941 v: 21859
w: 44509 v: 31459
w: 50011 v: 35351
w: 57637 v: 40745
w: 64763 v: 45785
w: 93179 v: 65881
w: 104701 v: 74029
w: 160421 v: 113431
w: 180259 v: 127459
w: 259363 v: 183395
w: 291437 v: 206075
w: 335891 v: 237509
w: 377429 v: 266881
w: 543061 v: 384001
w: 610219 v: 431489
w: 934987 v: 661135
w: 1050613 v: 742895
w: 1511669 v: 1068911
w: 1698611 v: 1201099
w: 1957709 v: 1384309
w: 2199811 v: 1555501
w: 3165187 v: 2238125
w: 3556613 v: 2514905
w: 5449501 v: 3853379
w: 6123419 v: 4329911
w: 8810651 v: 6230071
w: 9900229 v: 7000519
w: 11410363 v: 8068345
w: 12821437 v: 9066125
Sun Jan 27 18:51:33 PST 2019
w^2 - 2 v^2 = 1713719 = 7 17 14401
jagy@phobeusjunior:~$
answered Jan 28 at 2:53
Will JagyWill Jagy
104k5102201
$endgroup$
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
$$ w^2 - 2 v^2 = 7 cdot 17 cdot 14401 = 1713719$$
jagy@phobeusjunior:~$ date
Sun Jan 27 18:51:05 PST 2019
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 = 1713719 = 7 17 14401
Sun Jan 27 18:51:08 PST 2019
w: 1339 v: 199 SEED KEEP +-
w: 1381 v: 311 SEED KEEP +-
w: 1637 v: 695 SEED KEEP +-
w: 1763 v: 835 SEED KEEP +-
w: 1949 v: 1021 SEED BACK ONE STEP 1763 , -835
w: 2131 v: 1189 SEED BACK ONE STEP 1637 , -695
w: 2899 v: 1829 SEED BACK ONE STEP 1381 , -311
w: 3221 v: 2081 SEED BACK ONE STEP 1339 , -199
w: 4813 v: 3275
w: 5387 v: 3695
w: 7691 v: 5359
w: 8629 v: 6031
w: 9931 v: 6961
w: 11149 v: 7829
w: 16013 v: 11285
w: 17987 v: 12685
w: 27539 v: 19451
w: 30941 v: 21859
w: 44509 v: 31459
w: 50011 v: 35351
w: 57637 v: 40745
w: 64763 v: 45785
w: 93179 v: 65881
w: 104701 v: 74029
w: 160421 v: 113431
w: 180259 v: 127459
w: 259363 v: 183395
w: 291437 v: 206075
w: 335891 v: 237509
w: 377429 v: 266881
w: 543061 v: 384001
w: 610219 v: 431489
w: 934987 v: 661135
w: 1050613 v: 742895
w: 1511669 v: 1068911
w: 1698611 v: 1201099
w: 1957709 v: 1384309
w: 2199811 v: 1555501
w: 3165187 v: 2238125
w: 3556613 v: 2514905
w: 5449501 v: 3853379
w: 6123419 v: 4329911
w: 8810651 v: 6230071
w: 9900229 v: 7000519
w: 11410363 v: 8068345
w: 12821437 v: 9066125
Sun Jan 27 18:51:33 PST 2019
w^2 - 2 v^2 = 1713719 = 7 17 14401
jagy@phobeusjunior:~$
answered Jan 28 at 2:53
Will JagyWill Jagy
104k5102201
answered Jan 28 at 2:53
Will JagyWill Jagy
104k5102201
answered Jan 28 at 2:53
Will JagyWill Jagy
104k5102201
answered Jan 28 at 2:53
Will JagyWill Jagy
104k5102201
104k5102201
$begingroup$
I will work with what you have provided. I will be back.
$endgroup$
– ersh
Jan 28 at 2:58
1
$begingroup$
@ersh after $5,$ the next few $y$ that work are $11,15,21.$ Also, $y=40$ works, so some even $y$ are possible.
$endgroup$
– Will Jagy
Jan 28 at 3:21
add a comment |
$begingroup$
I will work with what you have provided. I will be back.
$endgroup$
– ersh
Jan 28 at 2:58
1
$begingroup$
@ersh after $5,$ the next few $y$ that work are $11,15,21.$ Also, $y=40$ works, so some even $y$ are possible.
$endgroup$
– Will Jagy
Jan 28 at 3:21
$begingroup$
I will work with what you have provided. I will be back.
$endgroup$
– ersh
Jan 28 at 2:58
$begingroup$
I will work with what you have provided. I will be back.
$endgroup$
– ersh
Jan 28 at 2:58
1
1
$begingroup$
@ersh after $5,$ the next few $y$ that work are $11,15,21.$ Also, $y=40$ works, so some even $y$ are possible.
$endgroup$
– Will Jagy
Jan 28 at 3:21
$begingroup$
@ersh after $5,$ the next few $y$ that work are $11,15,21.$ Also, $y=40$ works, so some even $y$ are possible.
$endgroup$
– Will Jagy
Jan 28 at 3:21
add a comment |
2
$begingroup$
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
answered Jan 28 at 1:34
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
2
$begingroup$
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
answered Jan 28 at 1:34
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
2
2
2
$begingroup$
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
answered Jan 28 at 1:34
Will JagyWill Jagy
104k5102201
$endgroup$
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
answered Jan 28 at 1:34
Will JagyWill Jagy
104k5102201
edited Jan 28 at 3:07
answered Jan 28 at 1:34
Will JagyWill Jagy
104k5102201
answered Jan 28 at 1:34
Will JagyWill Jagy
104k5102201
answered Jan 28 at 1:34
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
1
$begingroup$
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:~$
answered Jan 28 at 2:24
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
1
$begingroup$
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:~$
answered Jan 28 at 2:24
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
1
1
1
$begingroup$
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:~$
answered Jan 28 at 2:24
Will JagyWill Jagy
104k5102201
$endgroup$
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:~$
answered Jan 28 at 2:24
Will JagyWill Jagy
104k5102201
edited Jan 28 at 2:32
answered Jan 28 at 2:24
Will JagyWill Jagy
104k5102201
answered Jan 28 at 2:24
Will JagyWill Jagy
104k5102201
answered Jan 28 at 2:24
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
1
$begingroup$
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
answered Jan 28 at 3:36
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
1
$begingroup$
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
answered Jan 28 at 3:36
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
1
1
1
$begingroup$
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
answered Jan 28 at 3:36
Will JagyWill Jagy
104k5102201
$endgroup$
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
answered Jan 28 at 3:36
Will JagyWill Jagy
104k5102201
answered Jan 28 at 3:36
Will JagyWill Jagy
104k5102201
answered Jan 28 at 3:36
Will JagyWill Jagy
104k5102201
answered Jan 28 at 3:36
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
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$begingroup$
$(x,y,z)=(7,1,10)$ works (by a simple search).
$endgroup$
– lulu
Jan 28 at 0:20
$begingroup$
Good! Though I am expecting some parametric solution.
$endgroup$
– ersh
Jan 28 at 0:23
$begingroup$
Then you should edit your post to ask for that.
$endgroup$
– lulu
Jan 28 at 0:23
$begingroup$
Thanks!. Done!.
$endgroup$
– ersh
Jan 28 at 0:25