Possible bug in Solve function?
Clash Royale CLAN TAG#URR8PPP
Bug introduced in 10.0 and persisting through 11.3 or later
In 11.3.0 for Microsoft Windows (64bit) (March 7, 2018)
writing:
f[w_, x_, y_, z_] := w*x^2*y^3  z*(w^2 + x^2 + y^2  1)
eqn = {D[f[w, x, y, z], w] == 0,
D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0,
D[f[w, x, y, z], z] == 0};
sol = Solve[eqn];
Table[eqn /. sol[[n]], {n, Length[sol]}]
I get:
{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}
from which there are four wrong solutions.
Am I wrong or is it a Solve
bug?
EDIT: through the email address support@wolfram.com I contacted Wolfram Technical Support who in less than three working days have confirmed that it is a bug and have already proceeded to report to their developers.
add a comment 
Bug introduced in 10.0 and persisting through 11.3 or later
In 11.3.0 for Microsoft Windows (64bit) (March 7, 2018)
writing:
f[w_, x_, y_, z_] := w*x^2*y^3  z*(w^2 + x^2 + y^2  1)
eqn = {D[f[w, x, y, z], w] == 0,
D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0,
D[f[w, x, y, z], z] == 0};
sol = Solve[eqn];
Table[eqn /. sol[[n]], {n, Length[sol]}]
I get:
{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}
from which there are four wrong solutions.
Am I wrong or is it a Solve
bug?
EDIT: through the email address support@wolfram.com I contacted Wolfram Technical Support who in less than three working days have confirmed that it is a bug and have already proceeded to report to their developers.

1
You could use
List@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of>
.– Szabolcs
Jan 5 at 22:30 
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53 
5
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computerless♦
Jan 6 at 2:45
add a comment 
Bug introduced in 10.0 and persisting through 11.3 or later
In 11.3.0 for Microsoft Windows (64bit) (March 7, 2018)
writing:
f[w_, x_, y_, z_] := w*x^2*y^3  z*(w^2 + x^2 + y^2  1)
eqn = {D[f[w, x, y, z], w] == 0,
D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0,
D[f[w, x, y, z], z] == 0};
sol = Solve[eqn];
Table[eqn /. sol[[n]], {n, Length[sol]}]
I get:
{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}
from which there are four wrong solutions.
Am I wrong or is it a Solve
bug?
EDIT: through the email address support@wolfram.com I contacted Wolfram Technical Support who in less than three working days have confirmed that it is a bug and have already proceeded to report to their developers.
Bug introduced in 10.0 and persisting through 11.3 or later
In 11.3.0 for Microsoft Windows (64bit) (March 7, 2018)
writing:
f[w_, x_, y_, z_] := w*x^2*y^3  z*(w^2 + x^2 + y^2  1)
eqn = {D[f[w, x, y, z], w] == 0,
D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0,
D[f[w, x, y, z], z] == 0};
sol = Solve[eqn];
Table[eqn /. sol[[n]], {n, Length[sol]}]
I get:
{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}
from which there are four wrong solutions.
Am I wrong or is it a Solve
bug?
EDIT: through the email address support@wolfram.com I contacted Wolfram Technical Support who in less than three working days have confirmed that it is a bug and have already proceeded to report to their developers.

1
You could use
List@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of>
.– Szabolcs
Jan 5 at 22:30 
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53 
5
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computerless♦
Jan 6 at 2:45
add a comment 

1
You could use
List@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of>
.– Szabolcs
Jan 5 at 22:30 
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53 
5
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computerless♦
Jan 6 at 2:45
You could use List@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of >
.
– Szabolcs
Jan 5 at 22:30
You could use List@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of >
.
– Szabolcs
Jan 5 at 22:30
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computerless♦
Jan 6 at 2:45
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computerless♦
Jan 6 at 2:45
add a comment 
2 Answers
2
active
oldest
votes
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click “Help”, then “Give Feedback…”, and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.

Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11 
2
@TeM You should report it to Wolfram first, otherwise there’s no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31 
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44 
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment 
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3  z*(w^2 + x^2 + y^2  1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution > True]
$left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land y=0land
w=sqrt{1x^2}right)lor left(z=0land y=0land w=sqrt{1x^2}right)lor (z=0land y=1land x=0land w=0)lor (z=0land y=0land x=0land
w=1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land
x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}

Yes, of course, I had already tried. I was almost certain it was a bug from
Solve
, so I pointed out. Thank you very much anyway, always very kind!– TeM
Jan 6 at 1:14 
4
I believe
Solve
use the functionReduce
under the hood. when you removeBacksubstitution > True
, you’ll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..– Okkes Dulgerci
Jan 6 at 1:18
add a comment 
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click “Help”, then “Give Feedback…”, and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.

Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11 
2
@TeM You should report it to Wolfram first, otherwise there’s no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31 
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44 
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment 
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click “Help”, then “Give Feedback…”, and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.

Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11 
2
@TeM You should report it to Wolfram first, otherwise there’s no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31 
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44 
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment 
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click “Help”, then “Give Feedback…”, and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.
Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click “Help”, then “Give Feedback…”, and then fill out the form in your browser to report.
As another answer notes, you can always try Reduce
instead which may give better results in some cases, but Solve
is usually what you want.

Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11 
2
@TeM You should report it to Wolfram first, otherwise there’s no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31 
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44 
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment 

Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11 
2
@TeM You should report it to Wolfram first, otherwise there’s no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31 
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44 
4
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11
Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
Jan 5 at 22:11
@TeM You should report it to Wolfram first, otherwise there’s no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31
@TeM You should report it to Wolfram first, otherwise there’s no chance for it to get fixed.
– Szabolcs
Jan 5 at 22:31
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44
@Szabolcs: Could you direct me where I can do it correctly?
– TeM
Jan 5 at 22:44
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
@TeM wolfram.com/support/contact
– Szabolcs
Jan 5 at 23:49
add a comment 
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3  z*(w^2 + x^2 + y^2  1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution > True]
$left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land y=0land
w=sqrt{1x^2}right)lor left(z=0land y=0land w=sqrt{1x^2}right)lor (z=0land y=1land x=0land w=0)lor (z=0land y=0land x=0land
w=1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land
x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}

Yes, of course, I had already tried. I was almost certain it was a bug from
Solve
, so I pointed out. Thank you very much anyway, always very kind!– TeM
Jan 6 at 1:14 
4
I believe
Solve
use the functionReduce
under the hood. when you removeBacksubstitution > True
, you’ll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..– Okkes Dulgerci
Jan 6 at 1:18
add a comment 
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3  z*(w^2 + x^2 + y^2  1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution > True]
$left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land y=0land
w=sqrt{1x^2}right)lor left(z=0land y=0land w=sqrt{1x^2}right)lor (z=0land y=1land x=0land w=0)lor (z=0land y=0land x=0land
w=1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land
x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}

Yes, of course, I had already tried. I was almost certain it was a bug from
Solve
, so I pointed out. Thank you very much anyway, always very kind!– TeM
Jan 6 at 1:14 
4
I believe
Solve
use the functionReduce
under the hood. when you removeBacksubstitution > True
, you’ll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..– Okkes Dulgerci
Jan 6 at 1:18
add a comment 
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3  z*(w^2 + x^2 + y^2  1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution > True]
$left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land y=0land
w=sqrt{1x^2}right)lor left(z=0land y=0land w=sqrt{1x^2}right)lor (z=0land y=1land x=0land w=0)lor (z=0land y=0land x=0land
w=1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land
x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}
You can use Reduce
f[w_, x_, y_, z_] := w*x^2*y^3  z*(w^2 + x^2 + y^2  1)
eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0,
D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};
red = Reduce[eqn, Backsubstitution > True]
$left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land x=0land w=sqrt{1y^2}right)lor left(z=0land y=0land
w=sqrt{1x^2}right)lor left(z=0land y=0land w=sqrt{1x^2}right)lor (z=0land y=1land x=0land w=0)lor (z=0land y=0land x=0land
w=1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land
x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$
First@eqn //. {ToRules[red]}
{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}

Yes, of course, I had already tried. I was almost certain it was a bug from
Solve
, so I pointed out. Thank you very much anyway, always very kind!– TeM
Jan 6 at 1:14 
4
I believe
Solve
use the functionReduce
under the hood. when you removeBacksubstitution > True
, you’ll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..– Okkes Dulgerci
Jan 6 at 1:18
add a comment 

Yes, of course, I had already tried. I was almost certain it was a bug from
Solve
, so I pointed out. Thank you very much anyway, always very kind!– TeM
Jan 6 at 1:14 
4
I believe
Solve
use the functionReduce
under the hood. when you removeBacksubstitution > True
, you’ll find implicit solution, somehowSolve
messes up somewhere and it is definitely a bug..– Okkes Dulgerci
Jan 6 at 1:18
Yes, of course, I had already tried. I was almost certain it was a bug from Solve
, so I pointed out. Thank you very much anyway, always very kind!
– TeM
Jan 6 at 1:14
Yes, of course, I had already tried. I was almost certain it was a bug from Solve
, so I pointed out. Thank you very much anyway, always very kind!
– TeM
Jan 6 at 1:14
I believe Solve
use the function Reduce
under the hood. when you remove Backsubstitution > True
, you’ll find implicit solution, somehow Solve
messes up somewhere and it is definitely a bug..
– Okkes Dulgerci
Jan 6 at 1:18
I believe Solve
use the function Reduce
under the hood. when you remove Backsubstitution > True
, you’ll find implicit solution, somehow Solve
messes up somewhere and it is definitely a bug..
– Okkes Dulgerci
Jan 6 at 1:18
add a comment 
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You could use
List@ToRules@Reduce[eqn, {x, y, z, w}]
to get all valid solutions. Filter for those that only have numeric values on the RHS of>
.– Szabolcs
Jan 5 at 22:30
Select[sol, And @@ eqn /. # &]
– Bob Hanlon
Jan 5 at 23:53
Next time, please do not add the bugs tag yourself on a question. The tag is only supposed to be added after your observations have been confirmed by other users.
– J. M. is computerless♦
Jan 6 at 2:45