Polyomials with Ratioal Roots that Differ by a No-zero Costat Philip Gibbs The problem of fidig two polyomials P(x) ad Q(x) of a give degree i a sigle variable x that have all ratioal roots ad differ by a o-zero costat is ivestigated. It is show that the problem reduces to cosiderig oly polyomials with iteger roots. The cases < are solved geerically. For = the case of polyomials whose roots come i pairs (a,-a) is solved. For = 5 a ifiite umber of iequivalet solutios are foud with the asatz P(x) = -Q(-x). For = 6 a ifiite umber of solutios are also foud. Fially for = 8 we fid solitary examples. This also solves the problem of fidig two polyomials of degree that fully factorise ito liear factors with iteger coefficiets such that the differece is oe. Geeralities Let P(x) ad Q(x) be two polyomials of degree such that all roots of P(x) ad Q(x) are ratioal. For which degrees ca we fid two such polyomials which differ oly by a o-zero costat idepedet of x? Sice all roots are ratioal we ca factorise both polyomials P x = a (x r i ) Q x = b (x s i ) Where r i ad s i are the ratioal roots. For > 0, if the differece P(x) Q(x) = c is a costat idepedet of x the we must have a = b ad without loss of geerality we ca assume a = b = 1. Coditios o the roots The coditio for a solutio ca be writte i terms of the roots r i k k = s i, for 1 k < r i s i Equivalece to problems over itegers Take N to be the multiple of the deomiators of all roots for both equatios the P x = N P x/n = (x Nr i ) So P (x) is a polyomial with iteger roots. Similarly for Q (x) = N Q(x/N). But P (x) - Q (x) = N c is costat. It follows that if we ca fid a solutio with ratioal roots the we ca also fid oe with
iteger roots. The coverse is trivial. Therefore we eed oly search for polyomials with iteger roots r i ad s i. This problem is also equivalet to fidig two polyomials of degree that ca be fully factorised ito liear factors with iteger coefficiets ad that differ by oe. If we have two such polyomials the they have ratioal roots. Coversely, if we have two polyomials P(x) ad Q(x) with iteger roots that differ by a costat P(x) Q(x) = c, the substitute x = x s 1 where s 1 is a root of Q(x) so that Q (x ) = Q(x + s 1 ) has a factor of x ad therefore the product of the roots of P (x ) = P(x + s 1 ) is c. Now make a secod substitutio to defie P (x ) = (1/c)P (cx ) ad Q (x ) = (1/c)Q (cx ). It ca be verified that P (x ) ad Q (x ) factorise ito liear factors with iteger coefficiets ad that P (x ) - Q (x ) = 1 No coicidet roots If a root r of P(x) were also a root of Q(x) the x-r would be a factor of P(x)-Q(x) which could the ot be costat. It follows that oe of the roots of P(x) ca coicide with roots of Q(x). Trasformatios ad dualities Give oe solutio as roots r i ad s i, aother ca be formed by traslatig usig a iteger costat traslatio r i = r i + t, s i = s i + t, or by multiplyig by a costat r i = k.r i, s i = k.s i. Solutios which differ by combiatios of such trasformatios will be regarded as equivalet. Whe a solutio is equivalet to itself uder a o-trivial trasformatio we call it self-dual. This ca happe i essetially two ways; (1) r i = t r j ad s i = t s j () r i = t s j By usig the trasformatios we ca assume t = 0 i either case. Self-dual type (1) ca oly arise for eve sice otherwise it require a zero root for both polyomials. It is the equivalet to P(x) = P(-x) ad Q(x) = Q(-x) Self-dual type () ca arise for odd or eve. We the get P(x) = (-1) Q(-x) Whe we impose self duality o the solutios we automatically fulfil about half of the required coditios o the roots. This ca help us i the search at higher values of. = Geeral case The quadratic case is simple to solve. Usig the trasformatios we ca assume that the roots are (r, -r) ad (s, -s) from polyomials P(x) = x -r ad Q(x) = x -s. This provides solutios whe r s. All other solutios are equivalet to oe of these.
= Type () self -daul The cubic case is more challegig, but it is ot difficult to fid some type () self-dual solutios with This automatically gives us that The remaiig requiremet is that r 1 = -s 1, r = -s, r = -s r 1 + r + r = s 1 + s + s r 1 + r + r = 0 with oe of the roots equal to zero. This is easy to satisfy, i.e. take r 1 = a-b, r = b-c, r = c-a, where a, b ad c are ay distict itegers. This icludes a ifiite umber of iequivalet solutios e.g. by fixig a ad b, the varyig c. Geeral case For = it is also possible to costruct a more complete solutio. To see this first set r 1 = -s 1 This ca be doe without loss of geerality sice ay solutio is equivalet to oe with this coditio. The we require just, r + r = s + s s 1 = r + r - s - s The first equatio is well kow with geeral solutio i four parameters a, b, c, d, based o complex umber orms is give by r = ab + cd, r = ad - bc, s = ab - cd, s = ad + bc, The we ca complete the solutio by solvig the secod equatio with s 1 = cd - bc, r 1 = bc - cd To esure that the differece i the polyomials is o-zero, we eed r 1 r r - s 1 s s = bcd(b-d)(a-c)(a+c) 0 Although this solutio is complete up to traslatios it does ot reflect the permutatio symmetries of the origial problem. To fid a more symmetric solutio, first examie the matrix of differeces Each of these compoets factorises as follows δ ij = r i - s j
= c(b d) b(c a) d(c + a) b(a + c) cd a c (b d) d(a c) a + c (d b) bc Reame the factors as follows The the matrix takes a more symmetric form p = -c, q = c a, t = c + a, u = d b, v = b, w = -d = pu qv tw tv pw qu qw tu pv With the extra coditios p + q + t = u + v + w = 0 A further observatio is that a solutio ca also be derived from this matrix form by takig r i = j δ ij, s j = δ ij i It ca the be verified that the required equatios for the roots are satisfied without the extra coditios. To esure that the differece is o-zero we require r 1 r r - s 1 s s = (p-q)(p-t)(q-t)(u-v)(v-w)(u-w) 0 = Type (1) self -daul For quartic polyomials it is possible to fid type (1) self-dual solutios usig The remaiig equality we eed to satisfy is Which is solved usig r = - r 1, r = - r, s = - s 1, s = - s r 1 + r = s 1 + s r 1 = ab + cd, r = ad - bc, s 1 = ab - cd, s = ad + bc, From this we ca geerate a ifiite umber of iequivalet solutios. Type () self -daul We ca also look for type () self-dual solutios usig This the requires s 1 = - r 1, s = - r, s = - r, s = - r
r 1 + r + r + r = 0 ad r 1 + r + r + r = 0 the secod equatio gives The usig the secod equatio we get But we also have (r 1 + r )(r 1 - r 1 r + r ) + (r + r )(r - r r + r ) = 0 r 1 - r 1 r + r = r - r r + r (r 1 + r ) = (r + r ) Ad combiig the differet quadratics we also get (r 1 - r ) = (r - r ) It quickly follows that the two polyomials must be equal, so o type () self-dual solutios are possible for =. Type () self daul I this case we set =5 The the remaiig equatios to solve are r 1 = -s 1, r = -s, r = -s, r = -s, r 5 = -s 5 r 1 + r + r + r + r 5 = 0 ad r 1 + r + r + r + r 5 = 0 This has a ifiite umber of iequivalet solutios e.g. from this sequece for z ay positive iteger. r 1 = 1, r = z + z, r = -(z + z + 1), r = -(z + z + ), r 5 = z + 5z + =6 Type (1) self -daul For degree six polyomials cosider type (1) self-dual solutios usig The the remaiig equatios to solve are r = - r 1, r 5 = - r, r 6 = - r, s = - s 1, s 5 = - s, s 6 = - s r 1 + r + r = s 1 + s + s ad r 1 + r + r = s 1 + s + s Agai this has a ifiite umber of iequivalet solutios e.g. from this sequece. r 1 = 0, r = r = z + z + 1, s 1 = z + 1, s = z + z, s = z + z + 1
=8 Type (1) self -daul For =8 a brute force umerical search has produced some example solutios the smallest of which is { r i } = {-,-,-1,-5,5,1,,} { s i } = {-5,-1,-16,-,,16,1,5} Fial Remarks Solutios seem to be reasoably abudat up to degree 8 but there is o obvious patter that allows us to fid geeral solutios for arbitrarily high degree. As icreases the umber of variables icreases at twice the rate of the costraits, but the costraits are of icreasigly high degree. It is therefore a iterestig questio as to whether solutios exist for all. Although this problem has bee studied here for its ow iterest it may have applicatios to other problems where it is required to fid systems of umbers with small differeces that have may factors.