an irreducible cubic?
in [Math]
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From: quasi on 3 Aug 2010 03:16 On Tue, 03 Aug 2010 00:27:59 -0500, 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. > >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. Ok, I see how to prove it now. The polynomial f(x) with r integral and nonzero can't be reducible. Assume instead there is a nonzero integer r for which f(x) is reducible. Then as I argued in my prior message (quoted above), the polynomial g(y) = y^3 - (r^2 + r + 1) y^2 - (r^4 - r) y + (r^6 + r^5 + r^4) must have an integer root (where now y is an indeterminate). But by direct substitution, g(r^2) = r^3 g(r^2 + 1) = -r^3 So g has a real but non-integral root strictly between r^2 and r^2+1. Also g(-r^2-1) = -4r^4 - 3r^3 - 6r^2 - 2r - 2 g(-r^2+1) = 4r^4 + r^3 - 2r^2 But g(-r^2-1) is negative (even for all real r). and g(-r^2+1) is positive (since |r| >= 1). Also, g(-r^2) = -r^3 which is nonzero. It follows that g has a real but non-integral root strictly between -r^2 -1 and -r^2+1. Hence the 3rd root must be the integral root. Let the 3 roots be y1, y2, y3, where r^2 < y1 < r^2 + 1 -r^2-1 < y2 < -r^2+1 and y3 is the integral root. Adding the above inequalities yields -1 < y1 + y2 < 2 But the sum of the roots is r^2 + r + 1. It follows that r^2 + r - 1 < y3 < r^2 + r + 2 Since y3 is an integer, we must have y3 = r^2 + r or y3 = r^2 + r + 1 By divisibility considerations, it would have to be the latter one, but forget divisibility -- we don't need it. Instead just substitute each of the above for y3 and solve for r. As expected, the resulting equation for r has no integral roots. Therefore f(x) is always irreducible. Note that no use was made of the fact, proved in prior message, that r^2 + r + 1 would have to be a perfect cube. For this problem, inequality considerations trumped divisibility. quasi
From: quasi on 3 Aug 2010 03:29 On Tue, 03 Aug 2010 02:16:14 -0500, quasi <quasi(a)null.set> wrote: >On Tue, 03 Aug 2010 00:27:59 -0500, 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. >> >>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. > >Ok, I see how to prove it now. > >The polynomial f(x) with r integral and nonzero can't be reducible. > >Assume instead there is a nonzero integer r for which f(x) is >reducible. > >Then as I argued in my prior message (quoted above), the polynomial > > g(y) = y^3 - (r^2 + r + 1) y^2 - (r^4 - r) y + (r^6 + r^5 + r^4) > >must have an integer root (where now y is an indeterminate). > >But by direct substitution, > > g(r^2) = r^3 > g(r^2 + 1) = -r^3 > >So g has a real but non-integral root strictly between > > r^2 and r^2+1. > >Also > > g(-r^2-1) = -4r^4 - 3r^3 - 6r^2 - 2r - 2 > g(-r^2+1) = 4r^4 + r^3 - 2r^2 > >But g(-r^2-1) is negative (even for all real r). > >and g(-r^2+1) is positive (since |r| >= 1). > >Also, g(-r^2) = -r^3 which is nonzero. > >It follows that g has a real but non-integral root strictly between > > -r^2 -1 and -r^2+1. > >Hence the 3rd root must be the integral root. > >Let the 3 roots be y1, y2, y3, where > > r^2 < y1 < r^2 + 1 > -r^2-1 < y2 < -r^2+1 > >and y3 is the integral root. > >Adding the above inequalities yields > > -1 < y1 + y2 < 2 > >But the sum of the roots is r^2 + r + 1. It follows that > > r^2 + r - 1 < y3 < r^2 + r + 2 > > Since y3 is an integer, we must have > > y3 = r^2 + r > >or > > y3 = r^2 + r + 1 > >By divisibility considerations, it would have to be the latter one, >but forget divisibility -- we don't need it. Instead just substitute >each of the above for y3 and solve for r. As expected, the resulting >equation for r has no integral roots. More precisely, no nonzero integral roots. In fact, g(r^2 + r) = -r^3 and g(r^2 + r + 1) = r^3 + r^2 + r >Therefore f(x) is always irreducible. > >Note that no use was made of the fact, proved in prior message, that >r^2 + r + 1 would have to be a perfect cube. > >For this problem, inequality considerations trumped divisibility. quasi
From: Rob Johnson on 3 Aug 2010 11:56
In article <rbisrael.20100802214432$3c1d(a)news.acm.uiuc.edu>, Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: >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 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 >> > > > irreduc> > > ible 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 >> > >> > 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. I haven't been able to make this work any better than the previous attempt, but if P(x) = r x^3 - (r^2 + r + 1)x^2 - (r^3 - 1)x + r^2(r^2 + r + 1) then P(x-r+1) = r x^3 - (4r^2 - 2r + 1) x^2 + (r - 1)(4r^2 - 2r + 1) x + r(4r^2 - 2r + 1) Thus, P(x-r+1) does not satisfy Eisenstein if 4r^2 - 2r + 1 is squareful. We get the same type of Pell equation (2c)^2 d^3 - (4r-1)^2 = 3 [1] Perhaps it might be possible to show that the same r cannot work for [1] and [2] (2a)^2 b^3 - (2r+1)^2 = 3 [2] at the same time. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |