an irreducible cubic?
in [Math]
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From: Kent Holing on 31 Jul 2010 04:10 Can anybody prove or disprove that the cubic r x^3 - (r^2 + r + 1)x^2 - (r^3 - 1)x + r^2(r^2 + r + 1) = 0 is irreducible for all integers r /= 0? Kent Holing, Norway
From: Gerry on 31 Jul 2010 08:26 On Jul 31, 10:10 pm, Kent Holing <K...(a)statoil.com> wrote: > Can anybody prove or disprove that the cubic > r x^3 - (r^2 + r + 1)x^2 - (r^3 - 1)x + r^2(r^2 + r + 1) = 0 is irreducible for all integers r /= 0? > Kent Holing, > Norway No doubt you have observed that unless r^2 + r + 1 is squareful this polynomial is irreducible by Eisenstein. -- GM
From: Kent Holing on 1 Aug 2010 03:46 I have tried to find an r /= 0 such that the cubic is reducible. Extensive computer search has not revealed any such r. Is it possible to prove that the cubic is irreducible for all r /= 0? ________________________________________
From: Robert Israel on 1 Aug 2010 15:57 Gerry <gerry(a)math.mq.edu.au> writes: > On Jul 31, 10:10=A0pm, Kent Holing <K...(a)statoil.com> wrote: > > Can anybody prove or disprove that the cubic > > r x^3 - (r^2 + r + 1)x^2 - (r^3 - 1)x + r^2(r^2 + r + 1) =3D 0 is > > irreduc= > ible for all integers r /=3D 0? > > Kent Holing, > > Norway > > No doubt you have observed that unless r^2 + r + 1 is squareful > this polynomial is irreducible by Eisenstein. > -- > GM And it happens to be irreducible for r=18 and r=88916, which are the first two positive integers r for which r^2 + r + 1 is squareful. What's the next r? This sequence doesn't seem to be in the OEIS. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Gerry Myerson on 1 Aug 2010 23:02
In article <rbisrael.20100801195326$60e4(a)news.acm.uiuc.edu>, Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: > Gerry <gerry(a)math.mq.edu.au> writes: > > > On Jul 31, 10:10=A0pm, Kent Holing <K...(a)statoil.com> wrote: > > > Can anybody prove or disprove that the cubic > > > r x^3 - (r^2 + r + 1)x^2 - (r^3 - 1)x + r^2(r^2 + r + 1) =3D 0 is > > > irreduc= > > ible for all integers r /=3D 0? > > > Kent Holing, > > > Norway > > > > No doubt you have observed that unless r^2 + r + 1 is squareful > > this polynomial is irreducible by Eisenstein. > > -- > > GM > > And it happens to be irreducible for r=18 and r=88916, which are the first two > positive integers r for which r^2 + r + 1 is squareful. What's the next r? > This sequence doesn't seem to be in the OEIS. Squareful numbers are of the form a^2 b^3. r^2 + r + 1 = a^2 b^3 is (2 r + 1)^2 - b^3 (2 a)^2 = -3, so it's Pellish - if it has one solution, it has infinitely many. r = 18 leads to b = 7, so we're interested in solutions to x^2 - 343 y^2 = -3 with x odd and y even. The next one (after x = 37, y = 2, which corresponds to r = 17) has about 20 digits and can be found using http://www.alpertron.com.ar/QUAD.HTM r = 88916 comes from b = 13 and gives another infinite family. None of this answers Robert's question, of course, as to what the next integer is, although it does give a bound. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email) |