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?
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 8:48f[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 10:09Reduce
to do, when I specify the domain. $\endgroup$ – halirutan♦ Jan 7 '14 at 12:51Reduce[output, {m, k}, Integers]
to the OP's output, the extraneous solution goes away. $\endgroup$ – Michael E2 Jan 7 '14 at 14:09