The title alludes to a similar title of the paper [3] by Grunewald and Segal, in which it is shown how to solve a quadratic equation in integers. This latter procedure seems to be quite difficult, and the algorithm outlined in [3] is rather involved, although it is completely effective in the logical sense. 1991 Mathematics Subject Classification 11D0

Masser, D. W.

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HOW TO SOLVE A QUADRATIC EQUATIONIN RATIONALSD. W. MASSER1. IntroductionThe title alludes to a similar title of the paper [3] by Grunewald and Segal, inwhich it is shown how to solve a quadratic equation in integers. This latter procedureseems to be quite difficult, and the algorithm outlined in [3] is rather involved,although it is completely effective in the logical sense.Kornhauser has given fairly good explicit ‘ search bounds’ for integral solutionsif the number of variables is two or essentially at least five. Thus let P(X",…,Xn) bea polynomial of total degree at most 2, with rational integer coefficients, and writeH¯H(P) for the maximum of the absolute values of the coefficients.For the case n¯ 2, it is shown in [4] that if the equationP(X",X#)¯ 0 (1)has a solution in rational integers X"¯x", X#¯x#, then it has one withmax ²rx"r, rx#r´% (14H )&H. (2)For the case n& 5, assuming that the quadratic homogeneous part of P is non-singular, it is shown in [5] that if the equationP(X",…,Xn)¯ 0 (n)has a solution in rational integers X"¯x", …, Xn¯xn, then it has one withmax ²rx"r,…, rxnr´% (n$H )&!n(H­oH ). (3)Both (2) and (3) are actually close to best possible in their dependence on H :consideration of a generalized Pell-type equation shows that any bound exp ( f(H )) in(2) with limH!¢f(H )}H¯ 0 cannot be valid, and variations of a counterexample ofKneser show that any bound like (3) with exponent less than n}2 also cannot be valid.As far as I know, there are no explicit search bounds at all in the literature for thecases n¯ 3, n¯ 4 of three and four variables.However, if P happens to be homogeneous, then much more precise searchbounds are known, in this context for non-trivial solutions. These are traditionallystated for ‘classically integral ’ formsP(X",…,Xn)¯3ni="3nj="pijXiXjwith integers pij¯ pji. Thus the ‘off-diagonal ’ coefficients are even. In [1] (seeReceived 6 January 1997; revised 24 May 1997.1991 Mathematics Subject Classification 11D09.Bull. London Math. Soc. 30 (1998) 24–28how to solve a quadratic equation in rationals 25also [2, Lemma 8.1, p. 87]) Cassels proved that if (n) has an integral solutionX"¯x", …, Xn¯xnwith x",…,xnnot all zero, then there is one withmax ²rx"r,…, rxnr´% (3L)(n−")/#, (4)whereL¯L(P)¯3ni="3nj="rpijr.Furthermore, the original Kneser counterexample shows that the exponent (n®1)}2is best possible for every n. The result (4) has since been generalized in a number ofdirections, most recently by Schlickewei and Schmidt (see, for example, [6]).Now for rational solutions. If P remains homogeneous, then these are essentiallythe same as integral solutions, and so we shall regard P as not necessarilyhomogeneous. So it must now be written asP(X",…,Xn)¯3ni="3nj="pijXiXj­23ni="pi!Xi­p!!for integers pij¯ pji, and we defineL(P)¯3ni=!3nj=!rpijr.The Corollary of [3, p. 2] asserts the existence of an algorithm to decide if (n) has arational solution, via the Hasse–Minkowski Theorem. The main result of the presentpaper is a search bound of the same precision as Cassels’s estimate (4).Accordingly, for n¯ (n",…, nn) in Qn, defineH(n )¯0vmax ²1, rn"rv,…, rnnrv´,where the product is taken over all standard valuations  of the rational field Q ; if wewrite n"¯x"}x!, …, nn¯xn}x!for coprime integers x!1 0, x",…,xn, thenH(n )¯max ²rx!r, rx"r,…, rxnr´.This makes it clear that for any B there are only finitely many n in Qn withH(n )%B.Theorem. Suppose that (n) has a rational solution X"¯ n", …, Xn¯ nnforn¯ (n",…, nn). Then it has one withH(n )% (3L)(n+")/#.We shall also show that the exponent (n­1)}2 is best possible.Finally, in [3] an analogue for polynomials of the Hasse–Minkowski Theorem forforms was proved, namely the following.Theorem (Grunewald–Segal). Suppose that (n) has a solution oer the realnumbers R and oer each p-adic completion Qp. Then it has a solution oer Q.We shall give another short proof of this fact.26 d. w. masser2. ProofsIt is more convenient to work with forms, and it is clear that the above theoremon small rational zeros is equivalent to the following estimate for integral zeros.Proposition. Suppose that F(X!,…,Xn) is a classically integral quadratic form. Ifit has an integral zero (x!,…,xn) with x!1 0, then it has one withmax ²rx!r,…, rxnr´% (3L)(n+")/#,where L¯L(F ).As we said above, the exponent (n­1)}2 is best possible. It may be found slightlysurprising that the weak condition x!1 0 pushes up the ‘Cassels exponent’ (in thiscase n}2) by a positive amount.We prove the Proposition by induction on n. We start with n¯ 1 andF(X!,X")¯ aX #!­2bX!X"­cX#"for integers a, b, c withL¯ rar­2rbr­rcr.If c¯ 0, then b1 0 and now X!¯ 2b,X"¯®a is the required small zero. Otherwise,if c1 0, then the discriminant b#®ac must be the square of an integer d, andX!¯x!¯ c, X"¯x"¯®b­dis a zero. Further,4d #% 4rbr#­(rar­rcr)#%L#,and thereforerx!r%L, rx"r%L,as required.Now assume that n& 2 and that the Proposition has been proved for forms infewer than n­1 variables. To prove it for F(X!,…,Xn), we first apply Cassels’s resultto find a small integral zero (x!,…,xn), with x!,…,xnnot all zero, satisfyingmax ²rx!r,…, rxnr´% (3L)n/#.If x!1 0, then we are finished.Otherwise, if x!¯ 0, then we can find an even smaller zero. For then the formF(0,X",…,Xn) has a non-trivial integral zero, and now Cassels’s result with onefewer variable supplies integers x!",…,x!n, not all zero, withF(0,x!",…,x!n)¯ 0, M% (3L)(n−")/# (5)forM¯max ²rx!"r,…, rx!nr´.Assume now that the point x«¯ (x!!,x!",…,x!n) (with x!!¯ 0) is non-singular on thevariety V defined by F(X )¯ 0 for X¯ (X!,…,Xn). This point is ‘at infinity’, and weattempt to shift away from infinity by drawing chords. Take a generic point t, and letthe chord joining x« to t cut V in a new point. This new point is easily calculated(compare [2, p. 88]) to be proportional tox§¯F(t, t)x«®2F(t,x «) t, (6)where F(X,X «) is the bilinear form attached to F, with F(X,X )¯F(X ).how to solve a quadratic equation in rationals 27In other words, x§ is a zero of F for every t. We can now choose t¯ (t!,…, tn) sothat the coordinate x"!of x§¯ (x"!,…,x"n) is non-zero. Indeed,x"!¯®2F(t,x«) t!has the shape t!(f!t!­…­fntn) for f!,…, fnindependent of t ; and x« being non-singular means that f!,…, fnare not all zero. Now it is easy to find t!,…, tn, each either0 or 1, such that x"!1 0 as desired. (For example, we should take t!¯ 1; this sufficesif f"¯…¯ fn¯ 0, and otherwise, if say f"1 0, then either t"¯ 0 or t"¯ 1 will do.)Next, the definitions of L and M lead torF(t, t)r%L, rF(t,x«)r%LM,so (6) givesrx"ir%Lrx!ir­2LM rtir% 3LM (0% i% n).Thus (5) yieldsmax ²rx"!r,…, rx"nr´% 3L(3L)(n−")/#¯ (3L)(n+")/#,and therefore our desired small zero of F, at least if x« is non-singular.If x« is singular, then we reduce to fewer variables and use induction. But thecoefficients must stay under control. To ensure this, we can assume without loss ofgenerality that x!n1 0. Now x« is independent of the first n standard unit vectorsu!,…, un−", so we can define new variables Y!,…,YnbyX¯ (X!,…,Xn)¯Y!u!­…­Yn−"un−"­Ynx«. (7)In particular, X!¯Y!. Because x« is singular, it follows thatF(X!,…,Xn)¯F(Y!u!­…­Yn−"un−")¯G(Y!,…,Yn−")is a form G in the fewer variables Y!,…,Yn−". Furthermore, its coefficients constitutea subset of the coefficients of F. The given integral zero of F with X!1 0 suppliesan integral zero of G with Y!1 0. So the induction hypothesis provides integersy!,…, yn−", with y!1 0, such that G(y!,…, yn−")¯ 0 andmax ²ry!r,…, ryn−"r´% (3L)n/#.Now the values Y!¯ y!, …, Yn−"¯ yn−", Yn¯ 0 in (7) lead to X!¯ y!, …, Xn−"¯ yn−",Xn¯ 0, so that (y!,…, yn−", 0) is our required small integral zero of F. Thiscompletes the proof of the Proposition.To see that the exponent (n­1)}2 is best possible, it is more convenient to returnto inhomogeneous polynomials as in the Theorem. Fix an integer q& 2, and considerP(X",…,Xn)¯ 2X"®2q#®(X#®qX")#®…®(Xn®qXn−")#,where the squared linear terms are absent if n¯ 1. This has rational and even integralzeros, for exampleX"¯ q#, X#¯ q$, …, Xn¯ qn+".We proceed to show that any rational or even real zero X"¯ n",…,Xn¯ nnis almostas big, and in factnn& "#qn+". (8)28 d. w. masserIt is certainly clear that n"& q#, so (8) is true if n¯ 1. If n& 2, then withgi¯ ni+"®qni, we havenn®qn−"n"¯ 3n−"i="qn−"−igi,so the Cauchy–Schwarz inequality leads to(nn®qn−"n")#%Q 3n−"i="g#i,withQ¯ 3n−"i="q#(n−"−i) % 2q#(n−#).Because P(n",…, nn)¯ 0, the right-hand side of the above is exactly2Q(n"®q#)% 2Qn".It follows thatnn& qn−"n"®2qn−#on"¯ qn−#on"(qon"®2),which is at least qn−"(q#®2)& "#qn+", as asserted in (8).In particular, H(n )& "#qn+" for any rational zero n. But we see easily thatL¯ 2­2q#­(n®1) (1­2q­q#)% 4nq#.Therefore H(n )& cL(n+")/# for some c" 0 independent of q, and the exponent(n­1)}2 is best possible, as claimed.Finally, the strategy used in the proof of our Theorem leads to a quick proof ofthe Hasse–Minkowski Theorem for quadratic polynomials ; again it is moreconvenient to consider the equivalent version for forms F with zeros away frominfinity X!¯ 0, as in the Proposition. If F has everywhere non-trivial local zeros, thenthe classical Hasse–Minkowski Theorem provides a non-trivial global zero x«. If x« isnot at infinity, then we are finished. If x« is at infinity, then we attempt to shift awayto x§, as in (6). We succeed if x« is non-singular ; otherwise, if x« is singular, we reduceto fewer variables and use induction. The induction start, n¯ 1, is also easy, becauseif the discriminant is a square in R and each Qp, then it is a square in Q (see, forexample, [2, Lemma 3.1, p. 78]).References1. J. W. S. Cassels, ‘Bounds for the least solutions of homogeneous quadratic equations ’, Proc.Cambridge Philos. Soc. 51 (1955) 262–264; 52 (1956) 604.2. J. W. S. Cassels, Rational quadratic forms (Academic Press, London, 1978).3. F. J. Grunewald and D. Segal, ‘How to solve a quadratic equation in integers ’, Math. Proc.Cambridge Philos. Soc. 89 (1981) 1–5.4. D. Kornhauser, ‘On the smallest solution to the general binary quadratic diophantine equation’,Acta Arith. 55 (1990) 83–94.5. D. Kornhauser, ‘On small solutions of the general non-singular quadratic diophantine equation infive or more unknowns’, Math. Proc. Cambridge Philos. Soc. 107 (1990) 197–211.6. H. P. Schlickewei and W. M. Schmidt, ‘ Isotrope Unterra$ ume rationaler quadratischer Formen’,Math. Z. 201 (1989) 191–208.Mathematisches InstitutUniversita$ t BaselRheinsprung 214051 BaselSwitzerland

How to Solve a Quadratic Equation In Rationals

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