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Bug introduced in 10.0 and persisting through 11.3 or later

In 11.3.0 for Microsoft Windows (64-bit) (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.

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    $\begingroup$ 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 ->. $\endgroup$ – Szabolcs Jan 5 '19 at 22:30
  • $\begingroup$ Select[sol, And @@ eqn /. # &] $\endgroup$ – Bob Hanlon Jan 5 '19 at 23:53
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    $\begingroup$ 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. $\endgroup$ – J. M.'s ennui♦ Jan 6 '19 at 2:45
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    $\begingroup$ In version 12.0 bug is fixed. $\endgroup$ – Mariusz Iwaniuk Apr 17 '19 at 20:14
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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.

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  • $\begingroup$ Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^ $\endgroup$ – TeM Jan 5 '19 at 22:11
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    $\begingroup$ @TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed. $\endgroup$ – Szabolcs Jan 5 '19 at 22:31
  • $\begingroup$ @Szabolcs: Could you direct me where I can do it correctly? $\endgroup$ – TeM Jan 5 '19 at 22:44
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    $\begingroup$ @TeM wolfram.com/support/contact $\endgroup$ – Szabolcs Jan 5 '19 at 23:49
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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]

$\left(z=0\land x=0\land w=-\sqrt{1-y^2}\right)\lor \left(z=0\land x=0\land w=\sqrt{1-y^2}\right)\lor \left(z=0\land y=0\land w=-\sqrt{1-x^2}\right)\lor \left(z=0\land y=0\land w=\sqrt{1-x^2}\right)\lor (z=0\land y=-1\land x=0\land w=0)\lor (z=0\land y=0\land x=0\land w=-1)\lor (z=0\land y=0\land x=0\land w=1)\lor (z=0\land y=1\land x=0\land w=0)\lor \left(z=-\frac{1}{4 \sqrt{3}}\land y=-\frac{1}{\sqrt{2}}\land x=-\frac{1}{\sqrt{3}}\land w=\frac{1}{\sqrt{6}}\right)\lor \left(z=-\frac{1}{4 \sqrt{3}}\land y=-\frac{1}{\sqrt{2}}\land x=\frac{1}{\sqrt{3}}\land w=\frac{1}{\sqrt{6}}\right)\lor \left(z=-\frac{1}{4 \sqrt{3}}\land y=\frac{1}{\sqrt{2}}\land x=-\frac{1}{\sqrt{3}}\land w=-\frac{1}{\sqrt{6}}\right)\lor \left(z=-\frac{1}{4 \sqrt{3}}\land y=\frac{1}{\sqrt{2}}\land x=\frac{1}{\sqrt{3}}\land w=-\frac{1}{\sqrt{6}}\right)\lor \left(z=\frac{1}{4 \sqrt{3}}\land y=-\frac{1}{\sqrt{2}}\land x=-\frac{1}{\sqrt{3}}\land w=-\frac{1}{\sqrt{6}}\right)\lor \left(z=\frac{1}{4 \sqrt{3}}\land y=-\frac{1}{\sqrt{2}}\land x=\frac{1}{\sqrt{3}}\land w=-\frac{1}{\sqrt{6}}\right)\lor \left(z=\frac{1}{4 \sqrt{3}}\land y=\frac{1}{\sqrt{2}}\land x=-\frac{1}{\sqrt{3}}\land w=\frac{1}{\sqrt{6}}\right)\\ \lor \left(z=\frac{1}{4 \sqrt{3}}\land y=\frac{1}{\sqrt{2}}\land x=\frac{1}{\sqrt{3}}\land w=\frac{1}{\sqrt{6}}\right)$

First@eqn //. {ToRules[red]}

{True, True, True, True, True, True, True, True, True, True, True, True, True, True, True, True}

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  • $\begingroup$ 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! $\endgroup$ – TeM Jan 6 '19 at 1:14
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    $\begingroup$ 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.. $\endgroup$ – OkkesDulgerci Jan 6 '19 at 1:18

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