Hilbert's Tenth Problem is Unsolvable Author(s): Martin Davis Source: The American Mathematical Monthly, Vol. 80, No. 3 (Mar., 1973), pp. 233-269 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2318447 . Accessed: 22/03/2013 11:53 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to The American Mathematical Monthly. http://www.jstor.org This content downloaded from 129.2.56.193 on Fri, 22 Mar 2013 11:53:28 AM All use subject to JSTOR Terms and Conditions HILBERT'S TENTH PROBLEM IS UNSOLVABLE MARTIN DAVIS, CourantInstitute of MathematicalScience Whena longoutstanding problem is finallysolved, every mathematician would like to sharein the pleasureof discoveryby followingfor himself what has been done.But too oftenhe is stymiedby the abstruiseness of so muchof contemporary mathematics.The recentnegative solution to Hilbert'stenth problem given by Matiyasevic(cf. [23], [24]) is a happycounterexample. In thisarticle, a complete accountof thissolution is given;the only knowledge a readerneeds to followthe argumentis a littlenumber theory: specifically basic information about divisibility of positiveintegers and linearcongruences. (The materialin Chapter1 and the firstthree sections of Chapter2 of [25] morethan suffices.) Hilbert'stenth problem is to givea computingalgorithm which will tell of a givenpolynomial Diophantine equation with integer coefficients whether or notit hasa solutioninintegers.Matiyasevic proved that there is nosuch algorithm. Hilbert'stenth problem is thetenth in thefamous list which Hilbert gave in his 1900address before the International Congress of Mathematicians(cf. [18]). The wayin whichthe problem has beenresolved is verymuch in thespirit of Hilbert's addressin whichhe spoke of the convictionamong mathematicians "that every definitemathematical problem must necessarily be susceptibleof a precisesettlement, eitherin theform.of an actualanswer to thequestion asked, or by theproof of the impossibilityof itssolution ..." (italics added). Concerningsuch impossibility proofs Hilbertcommented: "Sometimesit happensthat we seekthe solution under unsatisfied hypotheses or in an inappropriatesense and are thereforeunable to reachour goal. Then the task arisesof provingthe impossibilityof solvingthe problemunder the given hypothesesand in thesense required. Such impossibility proofs were already given bythe ancients, in showing,e.g., that the hypotenuse of an isoscelesright triangle has an irrationalratio to itsleg. In modernmathematics the question of the impos- sibilityof certainsolutions has playeda key role,so thatwe have acquiredthe knowledgethat such old and difficultproblems as to provethe parallel axiom, to squarethe circle, or to solveequations of the fifth degree in radicals have no solution in theoriginally intended sense, but nevertheless have been solved in a preciseand completelysatisfactory way." MartinDavis receivedhis Princeton Ph. D. underAlonzo Church. He has heldpositions at Univ. ofIllinois, IAS, Univ. of Calif.-Davis,Ohio StateUniv., Rensselaer Poly, Yeshiva Univ. and New York Univ.,and he spenta leave at WestfieldCollege, London. He has done researchin various aspectsof thefoundations of mathematics,and is the authorof Computabilityand Unsolvability (McGraw-Hill,1958), The Undecidable (editor, Raven Press, 1965), Lectures on ModernMathematics (Gordonand Breach,1967), and First Coursein FunctionalAnalysis (Gordon and Breach,1967). Editor. 233 This content downloaded from 129.2.56.193 on Fri, 22 Mar 2013 11:53:28 AM All use subject to JSTOR Terms and Conditions 234 MARTIN DAVIS [March Matiyasevic'snegative solution of Hilbert's tenth problem is ofjustthis character. It is nota solutionin Hilbert's"originally intended sense" but rather a "preciseand completelysatisfactory" proof that no suchsolution is possible.The methods needed to makeit possible to provethe non-existence ofalgorithms had notbeen developed in 1900.These methods are part of the theory of recursive (or computable)functions, developedby logicians much later ([6] is an expositionof recursivefunction theory). In thisarticle no previousknowledge of recursive function theory is assumed.The littlethat is neededis developedin thearticle itself. Whatwill be provedin thebody of thisarticle is thatno algorithmexists for testinga polynomialwith integer coefficients to determinewhether or not it has positiveinteger solutions (Hilbert inquired about arbitrary integer solutions). But thenit willfollow at once thatthere can be no algorithmfor integer solutions either.For one could testthe equation P(Xl, 'Xn) "'O forpossession of positivesolutions <x1, *,x,> bytesting 2 2 P(1 + p 2+ q1 + r2+ s1 ,1 +p +q2 + rn+ S2) = 0 forpossession of integer solutions <p1, q1, r1,s1, s * * Pn qn rn,Sn>. This is because(by a well-knowntheorem of Lagrange)every non-negative integer is the sum of four squares.(Just this once the stated prerequisite is exceeded!Cf. [17], p. 302.)In the body of thisarticle, only positive integers will be dealt with-except whenthe contraryis explicitlystated. WhenMatiyasevic announced his beautifuland ingenioussolution in January 1970,it had been known for a decadethat the unsolvability of Hilbert'stenth problem wouldfollow if one couldconstruct a Diophantineequation whose solutions were such thatone of its componentsgrew roughly exponentially with another of its components.(In ?9,this is explainedmore precisely.) Matiyasevic showed how the Fibonaccinumbers could be usedto constructsuch an equation.In thisarticle the historicaldevelopment of the subject will not be followed;the aim has rather been to giveas smoothand straightforward an account of themain results as seemscurrently feasible.A briefappendix gives the history. 1. DiophantineSets. In thisarticle the usual problem of Diophantineequations willbe inverted.Instead of beinggiven an equationand seekingits solutions,one willbegin with the set of "solutions" and seek a correspondingDiophantine equation. More precisely: DEFINITION. A setS ofordered n-tuples of positive integers is called Diophantine .. ifthere is a polynomialP(xl, ...,9xny1, Ym), wherem ? 0, withinteger coefficients suchthat a given*n-tuple <x1, * , xn> belongsto S ifand onlyif there exist positive integersYi 1*, ymfor which This content downloaded from 129.2.56.193 on Fri, 22 Mar 2013 11:53:28 AM All use subject to JSTOR Terms and Conditions 1973] HILBERT'S TENTH PROBLEM IS UNSOLVABLE 235 P(XI*j," ,Xnj,Yl *...sYm) = 0- Borrowingfrom logic the symbols"3" for"there exists" and ".*." for "if and onlyif", the relation between the set S andthe polynomial P can be written succinctly as: <XI, .* sXn>E S-(3yjj,-*,Ym) [P(xjj, * sXniY1i .., Ym)= O?]s or equivalently: S = {<xl **">Xn | (3 y, * ,Ym) [P(xl 31*XnsYlq .. 9Ym)-?]}*- Notethat P may(and in non-trivialcases always will) have negative coefficients. The word"polynomial" should always be so construedin thearticle except where thecontrary is explicitly stated. Also all numbersin this article are positiveintegers unlessthe contrary is stated. The mainquestion which will be discussed(and settled)in thisarticle is: Whichsets are Diophantine?A vagueparaphrase of theeventual answer is: any set whichcould possibly be Diophantineis Diophantine.What does the phrase "whichcould possibly be Diophantine" mean? And how is all thisrelated to Hilbert's tenthproblem? These quite reasonable questions will only be answeredmuch later. In themeantime, the task will be developingtechniques for showing that various sets are indeedDiophantine. A fewvery simple examples: (i) thenumbers which are notpowers of 2: xeS..(3 y,z)[x = y(2z + 1)], (ii) thecomposite numbers: xeS.S*(3y,z) [x = (y + 1)(z + 1)], (iii) theordering relation on the positive integers; that is thesets {<x, y> x < y}, {<x,y> Ix _y}: x < y (3 z) (x + z = y), x:!gy.(3z) (x +z- I =y), (iv) thedivisibility relation; that is {<x,y> I x I y}: xj y..(3 z) (xz = y). Examples(i) and(ii) suggest,as othersets to consider,the set of powers of 2 and ofprimes respectively. As weshall eventually see, these sets are Diophantine; but the proofis not at all easy. Anotherexample: (v) theset W of <x,y, z> forwhich xj y and x < z: Here xl y.*.(3u) (y = xu) and x <z.*.(3v) (z = x + v). This content downloaded from 129.2.56.193 on Fri, 22 Mar 2013 11:53:28 AM All use subject to JSTOR Terms and Conditions 236 MARTINDAVIS [March Hence, <x, y,z> EW (3it, v)[(y-xu)2 + (z-x-v)2 =O]. Notethat the technique just used is perfectly general. So, in defininga Diophantine set one mayuse a simultaneoussystem P1 = O, P2 = 0, *,Pk = 0 of polynomial equationssince this system can be replacedby the equivalent single equation: 2 P1 +P22k + o+ Pk2=0? Bya "function"a positiveinteger valued function of one or more positive integer argumentswill alwaysbe understood. DEFINITION.A functionf of n argumentsis called Diophantineif .. {<XI, xn,Y>l y =f(xi, ,xX) is a Diophantine set, (i.e., f is Diophantineif its "graph" is Diophantine). Anotherquestion that will be answeredhere is: