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Bug introduced in 6.0 and fixed in 10.0

I want to find the integer values $k$ so that D[f[x], x] has integer solutions. I tried 

f[x_] := 2 x^3 - 3 (m - 1) x^2 + k * (m + 2)*(m - 3) x + 1
g := D[f[x], x]
d = Discriminant[g, x]
Reduce[{d == 0, -10 <= k <= 10}, {k, m}, Integers]

I got

(m == -3 && k == 4) || (m == 1 && k == 0) || (m == 7 && k == 3/2)

As you can see in the last solution k=3/2 which is definitely not an integer, as I have specified in the domain used by Reduce.

Is this a bug? Can anyone comment on this?

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  • $\begingroup$ You appear to have two negative signs in your definition of f[x]. In any case, I tried both $(m-1)$ and $(m+1)$, copied directly from your code, and I did NOT get your results. What version are you using? I am using 9.0.1. Oh, and what was your value of the discriminant? I got $d = 6912000 (-1 + m)^3 (-6 k - k m + k m^2)^3$ for $(m-1)$. $\endgroup$ – heropup Jan 7 '14 at 8:48
  • $\begingroup$ I am not using 9.0.1 $\endgroup$ – minthao_2011 Jan 7 '14 at 10:04
  • $\begingroup$ I need more information. What is the correct form for f[x]? Which version of Mathematica are you running? What is the value of $d$ that you obtained from your code? If you want help, you need to tell us more. $\endgroup$ – heropup Jan 7 '14 at 10:09
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    $\begingroup$ @rasher I think this is only formulated badly in the documentation. The line directly above in the details section states Reduce[expr,vars,dom] restricts all variables and parameters to belong to the domain dom. which clearly is what I would expect Reduce to do, when I specify the domain. $\endgroup$ – halirutan♦ Jan 7 '14 at 12:51
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    $\begingroup$ If you apply Reduce[output, {m, k}, Integers] to the OP's output, the extraneous solution goes away. $\endgroup$ – Michael E2 Jan 7 '14 at 14:09
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This has been confirmed as a bug.

The following also gives non-integer solutions which is clearly incorrect.

Reduce[d == 0 && -10 <= k <= 10 && k \[Element] Integers && m \[Element] Integers, {k, m}]
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  • $\begingroup$ This is so that the question gets off the unanswered list and Community stops bumping it up. $\endgroup$ – Szabolcs Feb 7 '14 at 18:55
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not an answer but may aid the discussion, the problem arises with the more simple form:

p = RandomInteger[100];
q = RandomInteger[{2, 100}];
Reduce[{k == p/q ( m^2 - 2 m + 1)/(m^2 - m - 6) , m > 4 }, {k, m},   Integers]

Produces the anomolous result m==7,k==p/q for "many" values of p,q.

Anyone arguing that k is somehow not supposed to be restricted needs to explain why there is just the one odd result, where obviously there is a rational k for any integer m (except 3,-2 of course)..

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    $\begingroup$ No need for any dispute, it's been recognized as a bug. $\endgroup$ – Daniel Lichtblau Jan 8 '14 at 18:01

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