an irreducible cubic? [Math]

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From: Robert Israel on 2 Aug 2010 17:53

Gerry Myerson <gerry(a)maths.mq.edi.ai.i2u4email> writes:

> 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.
>

Trying all b <= 500, I get 228594845490 for b = 199, then 9494342272670 for
b=19, and 690845140450082 and 1262403975253755261 for b=7. Kent's polynomial
is irreducible for each of these r values.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

From: quasi on 3 Aug 2010 00:24

From: Gerry Myerson on 3 Aug 2010 00:15

In article <lk5f56tgkou9affkfr4e1bqfq6qiq3r2da(a)4ax.com>,
quasi <quasi(a)null.set> wrote:

> On Sat, 31 Jul 2010 05:26:14 -0700 (PDT), Gerry <gerry(a)math.mq.edu.au>
> wrote:
>
> >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.
>
> A note on terminology ...
>
> The above usage of "squareful" is the Wikipedia one:
>
> <http://en.wikipedia.org/wiki/Powerful_number>
>
> However Wolfram's MathWorld gives a completely different definition:
>
> <http://mathworld.wolfram.com/Squareful.html>
>
> I'm pretty sure MathWorld has it completely wrong.

Agreed. Hey, welcome back, quasi.

--
Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)

From: quasi on 3 Aug 2010 01:27

On Sat, 31 Jul 2010 05:26:14 -0700 (PDT), Gerry <gerry(a)math.mq.edu.au>
wrote:

>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.

Based on divisibility considerations, I think I've shown more --
namely that if a nonzero integer r is such that the polynomial in
question is reducible, then r^2 + r + 1 must be a perfect cube.

The argument goes as follows ...

Assume r is a nonzero integer such that the polynomial is reducible.

Let f(x) be the univariate polynomial in x for the given r. Since f is
cubic, the reducibility of f implies that f has a rational root. Since
the leading coefficient is r, the polynomial g(y) defined by

g(y) = r^2 f(y/r)

must have an integral root. Expanding g(y) yields

g(y) = y^3 - (r^2 + r + 1) y^2 - (r^4 - r) y + (r^6 + r^5 + r^4)

By slight abuse of notation, let y be the integral root. Thus

y^3 - (r^2 + r + 1) y^2 - (r^4 - r) y + (r^6 + r^5 + r^4) = 0.

Let p be any prime factor of r^2 + r + 1.

It's clear that r^2 + r + 1 divides y^3, so p|^3, hence p|y.

Suppose p^k || r^2 + r + 1 (which means k is the greatest positive
integer such that p^k | (r^2 + r + 1).

Consider the powers of p which divide each of the 4 terms of g(y).

p^a || y^3
p^b || (r^2 + r + 1) y^2
p^c || -(r^4 - r) y
p^k || (r^6 + r^5 + r^4)

Then k < c and k < b, hence we must have k = a.

But a is a multiple of 3, hence so is k.

Therefore r^2 + r + 1 is a perfect cube, as claimed.

quasi

From: quasi on 3 Aug 2010 01:31

On Tue, 03 Aug 2010 14:15:08 +1000, Gerry Myerson
<gerry(a)maths.mq.edi.ai.i2u4email> wrote:

>In article <lk5f56tgkou9affkfr4e1bqfq6qiq3r2da(a)4ax.com>,
> quasi <quasi(a)null.set> wrote:
>
>> On Sat, 31 Jul 2010 05:26:14 -0700 (PDT), Gerry <gerry(a)math.mq.edu.au>
>> wrote:
>>
>> >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.
>>
>> A note on terminology ...
>>
>> The above usage of "squareful" is the Wikipedia one:
>>
>> <http://en.wikipedia.org/wiki/Powerful_number>
>>
>> However Wolfram's MathWorld gives a completely different definition:
>>
>> <http://mathworld.wolfram.com/Squareful.html>
>>
>> I'm pretty sure MathWorld has it completely wrong.
>
>Agreed. Hey, welcome back, quasi.

Thanks.

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