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
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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 thepleasure of 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.
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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
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
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 writtensuccinctly as:
S = {
(iv) thedivisibility relation; that is {
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 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, .. . { T(n) = 1 + 2 + *. +n - (n + 1) 2 SinceT(n) is an increasingfunction, for each positiveinteger z, thereis a unique n ? 0 suchthat T(n) < z < T(n + 1) = T(n) +n+ 1. Hence each z is uniquely representableas: z=T(n)+y; y z = T(x + y -2) + y. In this case, one writesx = L(z), y = R(z); also one sets P(x,y) = T(x + y -2) + y - 1. Notethat L(z), R(z) and P(x,y) are Diophantinefunctions since z = P(x,y) 2z=(x+y-2)(x+y-1)+2y x = L(z) (3y)[2z=(x+y-2)(x+y-1)+2y] y = R(z) (3x) [2z = (x + y-2) (x + y-1) + 2y]. The functionP(x; y) mapsthe set of orderedpairs of positiveintegers one-one 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] HILBERTS TENTH PROBLEMIS UNSOLVABLE 237 ontothe set of positive integers. And, for each z, the ordered pair which is mappedinto z byP(x, y) is (L(z), R(z)). ("P" is for"pair", "L" for"left", and "R" for"right".) Note also thatL(z) < z, R(z) < z. To summarize: THEOREM 1.1 (PairingFunction Theorem'). There are Diophantinefunctions P(x,y), L(z), R(z) such that (1) for all x, y, L(P(x,y)) = x, R(P(x,y)) = y, and (2) for all z, P(L(z),R(z)) = z, L(z) < z, R(z) ? z. Anotheruseful Diophantine function isrelated to the Chinese Remainder Theorem, statedbelow: DEFINITION. The numbersmln **, mN arecalled an admissiblesequence of moduli if i #j impliesthat mi and mj are relativelyprime. THEOREM 1.2 (ChineseRemainder Theorem). Let a,, *,aN be any positive integersand let ml * mN be an admissiblesequence of moduli. Then thereis an x such that: x a, modml x a2 mod M2 x aN mod MN. TheChinese remainder theorem is provedfor example in [25], p. 33. (Thatx can be assumedpositive is notordinarily stated. But sincethe product of themoduli addedto a solutiongives another solution, this is obvious.) Now let the funetionS(i, u) be definedas follows: S(i, u) = wI wherew is theunique positive integer for which: w-L(u) mod 1 + iR(u) w < 1 + i R(u). Herew is simplythe least positive remainder when L(u) is dividedby 1 + i R(u). THEOREM 1.3 (Sequence Number Theorem). There is a Diophantinefunction S(i, u) such that (1) S(i, u) < u, and (2) for each sequencea,, ***,aN, thereis a numberu such that S(i,u) = a1 for 1 < i < N. Proof. The firsttask is to showthat S(i, u) as definedjust above, is a Diophantine 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 238 MARTIN DAVIS [March function.The claim is thatw S(i, u) ifand only if the following system of equations has a solution: 2u = (x+y-2)(x+y-1)+2y X w + z(1 + iy) 1+iy -w+v-1. This is because(by thediscussion leading to thePairing Function Theorem), the firstequation is equivalentto: x = L(u) and y = R(u). Then(using a techniquealready noted) one needs only sum the squares of the three equationsto see thatS(i, u) is Diophantine. Now S(i, u) < L(u) ? u. So finally,let al, *- aN be givennumbers. Choose y to be somenumber greater than each of at, * aN and divisibleby each of 1,2,**-, N. Thenthe numbers 1 + y, 1 + 2y,**, 1 + Ny are an admissiblesequence of moduli. (For, if dj 1 + iy and dfI +jy, i x3 a mod l+y x a2 mod l+ 2y x aN modl+Ny. Let u =P(x,y), so that x =L(u) and y = R(u). Then, for i =1,2, .. N a L=- L(u) mod 1 + iR(u) and ai < y = R(u) < I + iR(u). But thenby definition,ai = S(i, u). A strikingcharacterization of Diophantinesets of positiveintegers (cf. [26]) is given by: THEOREM 1.4. A set S ofpositive integers is Diophantineif and only if there isa polynomialP suchthat S isprecisely the set ofpositive integers in the rangeof P. Proof. If S is relatedto P(x1,** , xm)as in the theoremthen x E-S *(3 xl, ^* , xm)[x = p(xl , xm)]. Conversely,let PS(x,X1, XXm) x[ Q(x x1 Xm) Tn, i-XSX= Let P(x,xl ** * xm)= x[l _ Q2(s, XI ,.*.*, Xm)]. Then,i'f x E-S, chooseX,, ** >xmsuch 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 239 that Q(x,xl, ,Xm) = 0. Then P(x, xl, *' -,xm) = x; so x is in the rangeof P. On the otherhand, if z = P(x, x, *., Xm),z > 0, thenQ(x, xx, I xm)must vanish (otherwise 1-_Q2<_ 0) so thatz = x and x eS. 2. Twenty-foureasy lemmas. The first major task is to provethat the exponential functionh(n, k) = nk is Diophantine.This is thehardest thing we shallhave to do. The proofis in ?3. In thissection we developthe methods we shallneed, using the so-calledPell equation: X2 -dy2=1, X,Y>O, where M(* d =a21, a >1 Althoughthis is a famousequation with a considerableliterature,2 a self-contained treatmentis given.Note theobvious solutions to (*): x= 1 y=0 x=a y=l. LEMMA 2.1. There are no integersx , y, positive,negative, or zero, whichsatisfy (*)for which1 LEMMA 2.2. Let x,y and x',y' be integers,positive, negative, or zero which satisfy(*). Let X" + Y (x +yV) (x' + y' Vd). Then, x",y" satisfies(k). Proof.Taking conjugates:x" - y"Id = (x - y Id) (x' - y'Id). Multiplying gives: (x")2 - d(y")2 = (x2 - dy2) ((X')2 - d(y')2) = 1. DEFINITION. xn(a), y,(a) are definedfor n > 0, a > 1, by setting xn(a) + yn(a)/d = (a + Vd)n. Wherethe context permits, the dependence on a is not explicitlyshown, writing XnL .sYn( LEMMA2.3. xn,,Yn sati'sfy() 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 240 MARTIN DAVIS [March Proof.This follows at onceby induction using Lemma 2.2. LEMMA 2.4. Let x, y be a non-negativesolution of (*). Then for some n, x = xn, Y =Yn. Proof.To beginwith x + yVd > 1. Oj theother hand the sequence (a + I increasesto infinity.Hence for some n > 0O (a + Xn + Yn,fdi < x + yV/d < (xn + Yn,/d) (a + V-d). Since (xn + ynV/d) (xn- yn Jd) = 1, the number xn- yn /d is positive. H4ence, 1 < (x + yld) (Xn-Yn y,d) < a + >/d. But thiscontradicts Lemmas 2.1 and 2.2. The definingrelation: Xn+ Yn,Id = (a + >d)" is a formalanalogue of thefamiliar formula: (cosu) + (sinu)y1- 1 = eiU= (cos1 + (sin1)1 - 1)u, withxn playing the role of cos, Yn playing the role of sin and d playingthe role of -1. Thus,the familiar trigonometric identities have analogues in which-1 is replaced by d at appropriateplaces. For examplethe Pell equationitself x- dY2 = 1 is just the analogueof the Pythagoreanidentity. Next analoguesof the familiar additionformulas are obtained. LEMMA 2.5. xm?n= XmXn ? dYnymand Ym?n = XnYm? XmYn. Proof. Xm+n +Ym+nvJd = (a + Vd)m+n = (Xm + Ym V/d)(Xi + Yn>/d) - (Xmxn+ dYnYm) + (XnYm+ XmYn) Id. Hence, Xm+n = XmXn + dYnYm Ym+n = XnYm+ XmYn. Similarly,(Xm.n + Ym.n Id) (Xn+ YnId) = Xm+ ym-/d. So + Xm_n Ym-nn/d = (Xm + ym-/d) (xn d)- and one proceeds.as above. 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 241 LEMMA 2.6. Ym?i = a Ym? Xm,and xm?1 = ax, + dym. Proof.Take n = 1 in Lemma2.5. The familiarnotation (x, y) is usedto symbolizethe g.c.d. of x and y. LEMMA 2.7. (x.,,yR)= 1. Proof. IfdIx and djYn, then dfx 2-dy2 i.e., dj1. LEMMA 2.8. YnJYnk Proof.This is obviouswhen k = 1. Proceedingby induction, using the addition formula(Lemma 2.5), Yn(m+1) = XnYnm+ XnmYn By theinduction hypothesis Yn j Ynm. Hence, Yn IYn(m+ 1)- LEMMA 2.9. Yn|IYt if and onlyif nj t. Proof.Lemma 2.8 gives the implicationin one direction.For the converse supposeYnj Yt but ntt. So one can writet = nq + r, 0 < r < n. Then, Yt = XrYnq+ XnqYr. Since (by Lemma2.8) ynj Ynq,it followsthat Yn J XnqYr. But (yn,xnq)= 1- (If df Yn, d Xnq,then by Lemma 2.$ dfYnq which, by Lemma 2.7, implies d = 1.) Henceyn j Yr. But,since r < n,we haveYr < Yn(e.g., by Lemma 2.6). Thisis a contradiction. LEMMA 2.10. Ynk k xyn mod (yM,)3. Proof. Xnk+ Y,,kV/d (a + Idyk = (Xn+ YnQd)k ( x k-iy ?W nJdj12 j= nJ So, k Ynk = j(Jxkydjh-h)/2 j odd Butall termsof this expansion for which ]> I are O0 mod(yR)3. LEMMA 2.1 1. Yn I YnYn Proof. Set k = YR in Lemma 2.10. LEMMA2.12. If YnI Yf,then Yn t. 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 242 MARTIN DAVIS [March Proof. By Lemma 2.9, n I t. Set t = nk. Using Lemma 2.10, y2 k xn-7y., i.e., ynI kx'-71.But byLemma 2.7, (Yn, Xn) = 1. SO,Yn k and henceYn I t. LEMMA 2.13. xn+ = 2axn -xn_ and Yn+I = 2ayn-yn. Proof.By Lemma 2.6, Xn+ = axn + dyn, Ynl+ = aYn+ Xn Xn = aXn - dyn, Yn-1 = aYn - Xn, So, Xn+ Jr Xn-I = 2axw Yn+l + Yn-l = 2ayn. Thesesecond order difference equations, together with the initial values xo = 1, x1 = a, Yo = 0, Yi = 1, determinethe values of all the Xn,Yn. Various propertiesof thesesequences are easily established by checking them for n = 0, 1 and usingthese differenceequations to showthat the property for n + 1 can be inferredfrom its holdingfor n and n - 1. Somesimple (but important) examples follow: LEMMA2.14. yn-n mod a - 1. Proof.For n = 0,1 equalityholds. Proceeding inductively, using a -1, mod a-1: Yn+1- 2ayn-Yn-y 2n-(n-1) mod a-1. LEMMA2.15. If a _ b mod c, thenfor all n, xn(a)-x.(b), yn(a) yj(b) mod c. Proof.Again for n = 0,1 thecongruence is an equality.Proceeding by induction: yn+1(a) = 2ay,(a) - yl1(a) - 2by.(b)-y,,-1(b) mod c = 7,.+ Il(b). LEMMA2.16. Whenn is eveny,, is evenand whenn is odd y, is odd. Proof Yn+ = 2aYnYn - I Yn-, mod 2. So whenn is even,Yn Yo = 0 mod 2, and when n is odd, Yn-Yi = 1 mod 2. LEMMA 2.17. xn(a) - yn(a)(a - y) Y" mod 2ay _ y2 _ 1. Proof. xo - yo(a - y) = 1 and xl - yl(a - y) = y, so the resultholds for n1= 0 and 1. UsingLemma 2.13 and proceedingby induction: Xn+ - Yn+(a - y) = 2a[Xn - yn(a - Y)] - [Xn- Yn-1(a - y)] = 2ayn _ yn-I 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 243 = y"nI(2ay-1) = yn-I y2 yn+1 LEMMA 2.18. For all n, Yn+1 > Yn_ n. Proof. By Lemma 2.6, Yn+1 > Yn.Since yo = 0 > 0, it followsby inductionthat yn> n for all n. LEMMA 2.19. For all n, xn+1(a) > x"(a) > an; xn(a) < (2a)". Proof. By Lemmas 2.6 and 2.13 a x"(a) < xn+ 1(a) < (2a)xn(a). The result follows by induction. Next some periodicity properties of the sequence Xk are obtained. LEMMA 2.20. X2"n=- xj mod xn. Proof. By the addition formulas (Lemma 2.5) X2n j = Xnxn j + dYnYn j - dYn(YnXi ?XnYi) mod xn - dy 2xj mod xn = (x2- _)xj - - x; mod xn. LEMMA 2.21. X4"+J x; mod xn. Proof. By Lemma 2.20 x4n"j -X2n" ijxX mod xn. LEMMA 2.22. Let xi- xj mod xn, i Proof. First suppose xn is odd and let q = (xn - 1)/2. Then the numbers -q, -q+1, -q+2, ., -1,0 , 1, ,q - 1, q are a complete set of mutually incongruent residues modulo xn. Now by Lemma 2.19, 1 =xo < X < ... Using Lemma 2.6, x - I<- xn la < i xn; so xn_1 < q. Also by Lemma 2.20, the numbers Xn+1,Xn+29 ...X2n-1,X2n are congruent modulo xn respectively to: 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 244 MARTIN DAVIS [March Thusthe numbers x0, x1, x2, **. , x2 are mutuallyincongruent modulo xn. This gives the result. Nextsuppose xn is evenand let q = xn/2. In thiscase, it is thenumbers - q +1, - q +2, ... - 1,O,1, ... ~q - 1,q whichare a completeset of mutually incongruent residues modulo x". (For, - q = q mod xn.)As above,xn1 < q. So theresult will follow as above,unless xn-1 = q = x"/2, so thatx"+ --q mod x", in whichcase i = n-1, =n + 1 would contradictour result. But, by Lemma 2.6, xn= axn_l+ dYn- 1, so thatxn = 2xn_1implies a = 2 and yn-i= 0, i.e.,n = 1. So theresult can failonly fora =2, n = 1 and i =0, j =2. LEMMA 2.23. Let xj xi mod xn, n > O,O< i < n, O < j < 4n, then either j= i or j =4n-i. Proof. Firstsuppose j < 2n. Thenby Lemma2.22, = ia unlessthe exceptional case occurs.Since i > 0, thiscan onlyhappen if j = 0. Butthen i = 2 > 1 = n. Otherwise,let j > 2n and set j = 4n -j so 0 < j < 2n. By Lemma2.21, X _ Xj -xi modxn. Again j = i unlessthe exceptionalcase ofLemma 2.22 occurs. But this last is out of thequestion because i, j > 0. LEMMA 2.24. If O < i < n and xj xi mod xn, thenj +i mod4n. Proof. Write]= 4nq + j, 0 _ j < 4n. By Lemma2.21, X_Xj_X. mod xn. By Lemma2.23i= j or i = 4n - j. So, ]s j- +i mod4n. 3. Theexponential function. Consider the system of Diophantineequations: (I) x2 - (a2-1)y2 = 1 (II) u2 - (a2-1)v2 = 1 (III) s2 - (b2-1)t2 = 1 (IV) v = ry2 (V) b = 1+4py=a+qu (VI) s = x+cu (VII) t = k + 4(d-1)y 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 245 (VIII) y = k+e-1. Then it is possibleto prove: THEOREM 3.1. For given a, x, k, a > 1, thesystem I-VIII has a solutionin the remainingarguments y, u, v,s, t,b, r,p, q, c, d, e ifand onlyifx = xk(a). Proof.First let there be givena solutionof I-VIII. ByV, b > a > 1. ThenI, II, III imply(by Lemma2.4) thatthere are i, j, n > 0 suchthat x = xi(a), y = yi(a), u = x"(a), v = y(a), s = xj(b), t = yj(b). ByIV, y < v so thati < n. V and VI yieldthe congruences b-a mod x"(a); xj(b)-xi(a) mod x"(a) and by Lemma2.15 one gets also xj(b) xj(a) mod x"(a). Thus, xi(a) xj(a) mod x"(a). By Lemma2.24, (1) ij?+i mod 4n. Next,equation IV' yields (y, (a))' y|y(a). so thatby Lemma 2.12, yi(a)n 7 and (1) yields: (2) j -+ i mod 4yi(a). By equationV b -1 mod 4yi(a), so by Lemma2.14, (3) yj(b) j mod 4yi(a). By equationVII, (4) yj(b) k mod 4y,(a). Combining(2), (3), (4), (5) k-+ i mod 4y,(a). 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 246 MARTIN DAVIS [March Equation VIII yields k < yi(a) and by Lemma 2.18, i < yi(a). Since the numbers - 2y+ 1, -2y +2, , 1,0, 1, 2y forma complete set of mutuallyincongruent residues modulo 4y 4y,(a), these inequalitiesshow that (5) implies k = i. Hence x ix(a) = Xk(a). Conversely,let x = xk(a). Set y yk(a) so that I holds. Let m = 2kyk(a) and let u = x"(a), v =yr(a). Then II is satisfied.By Lemmas2.9 and 2.11 y2' v. Hence one can choose r satisfyingIV. Moreoverby Lemma 2.16, v is even so thatu is odd. By Lemma 2.7, (u,v) = 1. Hence (u, v 4y) = 1. (If p is a primedivisor of u and of4y, then p jy because u is odd, and hence p Iv since y v.) So by the Chinese Remainder Theorem(Theorem 1.2), one can findbo such that bo-1 mod 4y bo-a mod u. Since bo + 4juy will also satisfythese congruences, b, p, q satisfyingV can be found. III is satisfiedby settings = xk(b), t = yk(b). Since b > a, s = xk(b)> xk(a) = x. By Lemma 2.15 (usingV), s -x mod u. So c can be chosen to satisfyVI. By Lemma 2.18, t> k and by Lemma 2.14, t = k mod b-I and hence usingV, t = k mod 4y. So d can be chosen to satisfyVIT. By Lemma 2.18 again, y > k, so VIII can be satisfiedby settinge = y - k + 1. COROLLARY 3.2. The function g(z,k) = Xk(Z + 1) is Diophantine. Proof. Adjoin to the systemI-VIII: (A) a = z +1. By thetheorem, the system (A), 1-VIII has a solutionif and onlyif x = xk(a) = g(z, k). Thus a Diophantinedefinition of g can be obtainedin the usual way by summing the squares of 9 polynomials. Now at last it is possibleto prove: THEOREM 3.3. The exponentialfunctionh(n, k) =-nk is Diophantine. 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 247 First,a simpleinequality: LEMMA 3.4. If a > yk, then2ay - y- - 1 > yk. Proof.Set g(y) =2ay - y2 - 1. Then (since a _2) g(1) =2a-2 > a. For 1 IX (x-y(a-n)-m)2 =(f-1)2(2an-n2 - 1)2 X m + g = 2an-n2 2 XI w-n + h = k + l XII a2 -(w2-1) (w-1)2z2 = 1. Theorem3.3 then follows at oncefrom: LEMMA 3.5. m = nk ifand only if equationsI-XII have a solutionin the remain- ing argumen?ts. Proof. Suppose I-XII hold. By XI, w > 1. Hence (w - 1)z > 0 and so by XII a > 1. So Theorem3.1 appliesand it followsthat x = xk(a), y = yk(a). By IX and Lemma2.17, m-nk mod 2an-n2 1. XI yields k,n By XII (usingLemma 2.4),for some j, a = xj(w), (w - 1)z = yj(w). By Lemma 2.14, j 0 mod w - 1 so that j > w - 1. So by Lemma 2.19, a ? ww1 > n Now by X, m < 2an-n2 - 1, and by Lemma3.4, nk< 2an-n2 _ 1. Sincem and nk are congruentand bothless than the modulus, they must be equal. Conversely,suppose that m = nk. Solutionsmust be foundfor I-XII. Choose any numberw such that w > n and w > k. Set a = xw_I(w)so that a > 1. By Lemma 2.14, Yw-l(w) 0 mod w - 1. 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 248 MARTIN DAVIS [March So one can write Yw- (W) = z(w -1); thusXII is satisfied.XI can be satisfiedby setting h=w-n, I=w-k. As before,a > nk so thatagain by Lemma 3.4, m = nk < 2an - n 1 and X can be satisfied.Setting x = xk(a), y = yk(a), Lemma 2.17 permitsone to definef such that x - y(a - n) - m (f- 1)(2an -n2 so thatIX is satisfied.Finally, I-VIII can be satisfiedby Theorem 3.1. 4. Thelanguage of Diophantine predicates. Now thatit has beenproved that the exponentialfunctionis Diophantine, many other functions and sets can be handled. As an example,let h(u,v, w) = uv . The claim is thath is a Diophantinefunction. For: y uvW (3z) (y = z& vw) where"&" is the logician'ssymbol for "and". UsingTheorem 3.3, thereis a polynomialP such that: = y u4Zz~~~~~~~~~~~~ (3rj, *,rn)[P(y, u, z, rl, r 0,7n]9 = Z Vw 3S 1, *sSO [P(Z, VI ,Ws I SSn O= ] Then, y = uVW (3(z,ri,*,rn,si, * ,sn) [Pp2(y,u, z, ri,.*,r) + P2(Z, V,W, S1, *', S) = 0]. Now this procedureis perfectlygeneral: Expressionswhich are alreadyknown to yieldDiophantine sets may be combinedfreely using the logicaloperations of "&" and "(3)"; theresulting expression will again define a Diophantineset. (Such expressionsare sometimescalled Diophantinepredicates.) In this "language" it is also permissibleto use thelogician's "V" for"or", since: (3 rl, -,rn) [PI = O] V(OSI, ', Sm) EP2 = ?] 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 249 Threeimportant Diophantine functions are givenby: THEOREM 4.1. Thefollowing functions are Diophantine: (1) f(n,k) =( (2) g(n) = n! y (3) h(a, b,y) = H (a + bk). k=1 In provingthis theorem the familiar notation [oc], where a is a realnumber, will be usedto meanthe uniqueinteger such that [ac] ? a < Lx]+ 1. LEMMA 4.1. For 0 < k ! n, u > 2n [(u + 1)n/uk] = I', u i= k Proof. (u+1)n/uk= ?(I . u =S + R where n- R = ( .)uik S= jikk nI-3 '7 i- k ThenS is an integerand ' R < u (. i=0 i- Ivi < ui 1 ( i0O - u(I. + 1)n < 1. So, S ? (u + 1)n/uk < S + I whichgives the result. 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 250 MARTIN DAVIS [March LEMMA 4.2. For 0 < k < n, u > 2 , [(U + 1)nlUk]- kn odu Proof. In Lemma4.1 all termsof the sum for which i > k aredivisible by u. LEMMA 4.3. f(n, k) = () is Diophantine. Proof. Since (k--) =2n Lemma 4.2 determines(n) as the unique positive integercongruent to [(u + 1)n/uk]modulo u and < u. Thus, z-( (3u, v,w) (v = 2n& u > v k ) &w = [(u + I)n!uk] &z W mod u&z To see that(k) is Diophantine,it then sufficesto note that each of the above expressionsseparated by "&" areDiophantine predicates; v=2" isof course Diophan- tineby Theorem3. The inequalityu > v is of courseDiophantine since u > v (3x)(u = v + x). Also, z3w mod u & z w = [(u + 1)n /uk] (3x,y,t) (t = u + 1 &x = tn&y =u k&w ?xIy < w + 1), and w < x/y< w + 1 wy < x < (w + l)y. LEMMA 4.4. If r > (2x)x+1 then x= [r (rx)l] Proof. Let r > (2x)x+l. Then, X/ r ~~~rxx! x = r(r - 1) ..(r-x +1) = x! -, 4 \ , x 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 251 < xx x Now, 1 = 1 + x x 2 1 x--+ r r r~ ~~~ ll r 2 <1+-{1 + + 3: + + r r r 2x r And, j =l r 2xr~~~ +2x (x) 2x < 1+- 2 < I + 2x x r So, rxi(r) < x! + - .x!2x 2x+1 xx+' < x! + _ r < X! + 1. LEMMA 4.5. n! is a Diophantinefunction. Proof: m = n! * (3r,s, t,u, v) {s = 2x + 1 & t = x + 1 &r = St &u= r&v =(r) & mv < u < (m + 1)v}. n LEMMA 4.6. Let bq -a mod M. Then, I (a + bk) bYy!(y Y) mod M. 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 252 MARTINDAVIS [March Proof q + byY! = by(q+y) (q +y-l)-. (q+l1) = (bq + yb) (bq + (y-1)b) *.. (bq + b) 3 (a + yb) (a + (y-I)b) **(a + b) (mod M). LEMMA 4.7. h(a, b, y) = fkj= 1 (a + bk) is a Diophantine function. Proof. In Lemma 4.6 choose M = b(a + by)' + 1. Thein, (M, b) = 1 and M > fJ7k1 (a + bk). Hencethe congruence bq _ a modM is solvablefor q and then fk=1 (a + bk) is determinedas theunique number which is congruentmodulo M to byy! q + Y) and is also < M. I.e., y Z HI (a + bk) (M, p,q, r,s,t, u, v,w, x) k=1 r = a + by &s = rY & M = bs + 1 &bq a + Mt & u- by&v = y! &z < M & w =q?y &x= (w)&z+MP=uvx} Usingthe previousexpressions for the exponentialfunction, for v = y! and for x -(), we obtainthe result. The assertionof Theorem4.1 is containedin Lemmas4.3, 4.5, and 4.7. 5. Boundedquantifiers. The languageof Diophantinepredicates permits use of &, V, and 3. Otheroperations used by logicians are: for "not" (Vx) for "forall x" for "if*.., then ..." However,as willbe clearlater, the use of any of these other operations can lead to expressionswhich define sets that are notDiophantine. There are also thebounded existential quantifiers: '6(3y)s, " which means "(3y) (y < x & and thebounded universal quantifiers: "(Vy)x X whichmeans "(Vy) (y > xV P) 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 19731 HILBERT S TENTH PROBLEM IS UNSOLVABLE 253 It turnsout that these operations may be adjoinedto thelanguage of Diophantine predicates;that is, the setsdefined by expressionsof thisextended language will stillbe Diophantine.I.e., THEOREM 5.1. If P is a polynomial, R = { S { LEMMA 5.1. (Vk)gY(3y1, , Ym) [I(Y,k, XI, ,Xn,Y,, Ym) = 0] (3u) (Vk):,Y(3yI,*I* Ym) -u[P(y, k, xl1, ** , xn, Y I,* Ym)= ?] Proof.The fightside of theequivalence trivially implies the left side. For the converse,suppose the left side is true for given y, x1, * xn. Thenfor each k = 1,2, . ,y thereare definitenumbers (k)**, y(k) forwhich: P(y, k,x1 **,,(k) , , y(k)) 0. Takingu to be themaximum of the mynumbers {(k)| j = J, **,m; k = 1,2, **, y}, it followsthat the right side of theequivalence is likewisetrue. LEMMA 5.2. Let Q(y,Ux1 ...X) be a polynomialwith the properties: (1) ~~Q(Y'U, X1, *,Xn) > U, (2) Q(Y,U,X1, *" Xn)> Y, (3) k < y and yl, **,Ym _ u imply|P(y,k,xl-,*,xn,Yl, IYm)|, (Vk)5Sy(3y I *y,Y5)u [P(y, k, xl **"' XnsY, I ..* ~Ym)= 0] y (3c,t,a,, ...$am) [1 +ct= (1 + kt) k=1 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 254 MARTIN DAVIS [March u ... &t = QU3o8eX13 3,x)! & 1 + ct fI (a,l-j) j=i &... &1 +Ct f (amJ-(j) j =1 &P(y,c,-x,,xn,al, *,am)-=O mod 1+ct]. The pointof this lemma is thatwhile the right side of the equivalence seems the morecomplicated of the two, it is freeof bounded universal quantifiers. Proof.First the implication in the - direction: Foreach k = 1,2, **, y,let Pk be a primefactor of 1 + kt.Let y,k) be theremainder whena' is dividedby pk (k = 1,2,***,y; i = 1,2,**,m). It willfollow that for each k,i: (a) 1< Yk) < u (b) P(y,k,x, ...xnyk) mk)) - 0. Pk i + 1 1 To demonstrate(a),.~~~~~~~~~~~~~~~~~~~~~~~~~~~~note that kt, + kt| + ct and 1 + ct f[J>=1(a, -j). I.e., pkI1iJ=l(ai -j). Since Pk is a prime,Pkj ai -j forsome 1j2,j *.*?u. That is jaai= ) mod Pk. Since t = Q(y,u, x1, ,x)!, (2) implies that every divisor of 1 + kt must be .. > Q(y, u, xi, ** *, xn"). So Pk > Q(y,,u, xi, ** x") and by (1), pk > u. Hence j yik) = j To demonstrate(b), firstnote that 1+ct=1+kt=0 mod Pk. Hence k+kct-=c+kct mod Pk, i.e., k-c mod Pk. Wehave already obtained yk)-=ai mod p,. Thus, Y(f k,x1, * ",y)-P(y, c, xi , Xn.a,, X , am) 0 modpk. Finally 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 255 |P(y, k,xl, ...Y ,n,( (Y(ksx) )| _ This proves (b) and completesthe proofof the = implication. To prove the => implication,let P(Y'k> (k) ... Ymk)= ?~ for each k = 1,2, .* , t, whereeach yj(k)< u . We sett = Q(y,u,x1,..., x")!, and since Hk= (1 + kt) =1 mod t, we can findc suchthat y 1+ Ct = H (1 + kt). k= 1 Now, it is claimed that for 1 ? k < I< y, (1 + kt,1 + It) = 1. For, let p 11 + kt,p| l + It. Then P I-k, so p < y. But since Q(y,u,x1, ,xn) > y this impliesp I t whichis impossible.Thus the numbers1 + kt forman admissible sequence of moduli and the Chinese RemainderTheorem (Theorem 1.2) may be applied to yield,for each i, 1 < i < m, a numberai such that a, _yek) modI + kt, k-=1, 2, -*,y. As above, k _ c mod I + kt. So - P(y c,x -X,n aj, am) P(y, k,xl, **, X", k) ... y(k)) mod 1 + kt, - 0. Since the numbers 1 + kt are relativelyprime in pairs and each divides P(y, c, x1, ., x", a1, **, am) so does theirproduct. I.e., P(y, c, x xnax., 1, .. am) -0 mod 1 + ct. Finally, ai y=k) mod 1 + kt, i.e., 1 +kt az _ (k) Since1 u I+ ct Hrl (ai -j). j=1 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 256 MARTIN DAVIS [March Nowit is easyto completethe proof of Theorem 5.1 using Lemmas5.1 and 5.2. Firstfind a polynomialQ satisfying(1), (2), (3) of Lemma5.2. Thisis easyto do: Write N P(yuk, xj,. sX.,Yi ..sYm) Ya tr r=1 whereeach trhas theform S=|cy Xq,Xq,2. qnyl S2 .Sm tr=c'k 2 y 2 Y forc an integerpositive or negative.Set Ur = cya+bX 1XX2...xqnus1?S2+--+Sm and let N Q(Y' UsX,,.. sxn) = u + Y + E Ur. r=1 Then(1), (2), and (3) ofLemma 5.2 holdtrivially. Thus: (Vk) Y (3u, c, t,a,***,am) + ct ( + kt) k =1 &t=Q(y,uxI X ,x n)!&1 +ctI 171(a1 -j) j=1 u & ...&1 + ctlHi (a,n-]) j=l &P(Y'c,xj,-,x,,aj,-,am)--0 mod I + ct] (hu,c, t, a *a+ , ejh,g&1, *, gm,h1, ,e hn ) , Y &a t =f!&gD=iao-aU &e2=ya2-Uh &e&4gm.am1u u ti & h,= f (gI+ k) &h2 = fl (92 +k) k=1 k=1 t .. & &hm= rI (gmn+k) &e| hj&e| h2&}&e| hm k=1 & I= P(y' , xl,s,, x, a,,,,",an) &e 1 and this is Diophantine by Theorem 4.1. 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 257 6. Recursivefunctions. So far one trickafter another has been used to show thatvarious sets are Diophantine.But nowvery powerful methods are available:it turnsout thatthe expandedversion of the languageof Diophantinepredicates, permittingthe use ofbounded quantifiers (sanctioned by Theorem 5.1) together with theSequence Number Theorem (Theorem 1.3) enables one to show in quite a straight- forwardway that almost any set we pleaseis Diophantine. Someexamples are in order: (i) theset P of primenumbers: xeP x > I &(Vy,z) AnotherDiophantine definition of the primes is: X GP-=x > I &((x -1)! ,x) = I x > 1 &(3y,z,u,v) [y = x -1 &z = y! &(uz - vx)2 = 1]; butthe first definition is themore natural one. From Theorem 1.4 it followsthat thereis a "prime-representing"polynomial P, i.e.,a positiveinteger is primeif and onlyif it is in the rangeof P. For an explicitconstruction of such a polynomialP, cf. [23a]. (ii) thefunction g(y) = ,7ly=1 (I + k2). Here we use the SequenceNumber Theoremto "encode" the sequence g(1), g(2), . ,g(y) into a singlenumber u, i.e., so that S(i,u) = g(i), i = 1,2, .,y. Thus, z = g(y) >(3u) {S(1,u) = 2 &(Vk) (3u) {S(l,u) = 2 & (Vk)?y[k = 1 V (3a,b,c) (a k - 1 & b-S(a, u) & c = S(k, u) & c = (1 + k2)b)] & z S(y,-u)}. By now it is clear thatthe availablemethods are quite general.They are so powerfulthat the questionbecomes: how can any "reasonable"set or function escapethese methods, i.e., not be Diophantine? The strengthof the methods can be testedby considering the class ofall compu- tableor recursivefunctions. These are thefunctions which can be computedby a finiteprogram or computingmachine having arbitrarily large amounts of timeand memoryat its disposal. Many rigorous definitions ofthis class (all ofthem equivalent) are available.One of thesimplest is as follows: The recursive functions3 are all those functionsobtainable from the initial functions 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 258 MARTIN DAVIS [March c(x) = ,s(X) =x + 1; U"(x1',**x) i t < i COMPOSITION yieldsthe function , h(X 1, "XJ)fg, 1 (x 1,* ,X.),* g(x, ,X.)) fromthe given functionsg1, , gn and f(t1,* ,r, .) PRIMITIVE RECURSION yields the functionh(xl, **,x",z) whichsatisfies the equations: h(xl, x., t) =f(xl, **x.) h(xl,* * x" t + 1) = g(t,h(x1, * * *, Xn, t) xl, ** , xJ, romthe givenfunctions f, g. Whenn = 0,f becomesa constantso thath is obtaineddirectly from g. MINIMALIZATION yieldsthe function: h (x 1, *, xj= mi'n,U(x 1, *.*1 X,x,y) = g(X, X,XnY)] fromthe given functionsf, g assumingthatf, g are such that foreach x1, , x. there is at leastone y satisfyingthe equationf(x1, . ,x",y) = g(x1,.,x",y); (i.e., h must be everywheredefined). The mainresult of thisarticle is: TEILOREM6.1. AfunctionisDDiophantine if and only ifit is recursive. To beginwith, consider the following short list of recursive functions: (1) x + y is recursivesince x + I = s(x), x + (t + 1) = s(x + t) = g(t,x + t,x), whereg(u, V,w) = S(U3(U, V, W)). (2) x *y is recursivesince x l = U1(x) x (t+ 1)=(x t)+x=g(t,x tx), whereg(u, v,w) = U3(u, V,W) + U3(u, V,W). 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 259 (3) For each fixedk, the constant function Ck(X) = k is recursive,since c1(x) is one of the initialfunctions and Ck+I(X) = Ck(X) + c(x). (4) Any polynomialP(x1, ***, x") with positiveinteger coefficients is recursive, sinceany such function can be expressedby a finiteiteration of additionsand mul- tiplicationsof variablesand c(x). E.g., 2x2y + 3xz3 + 5 = C2(X) * X * X * y + C3(X) * X * Z * Z * Z + C5(X). So (1), (2), (3), and compositiongives the result. Now it is easyto see thatevery Diophantine function is recursive: Letf be Diophantine,and write: ... y =f(xj, *w,x.)<*(3tls *ss tm) [P(X1, ..,Xng Y9 tIg tm) .. = Q(XD, *' * Xn Y, tl, 9 tm)] 9 whereP, Q arepolynomials with positive integer coefficients. Then, by thesequence numbertheorem: f(x1, x..,x) = S(1, min,,[P(x1, X.,x",S(1, u), S(2, u), ... , S(m + 1,u)) = Q(xl, ..., x,, S(1, u), S(2, u), ..., S(m + 1,u))]). SinceP, Q, S(i, u) arerecursive, so isf (usingcomposition and minimalization). To obtainthe converse: S(i, u) is knownto be Diophantine;the otherinitial functionsare triviallyDiophantine. Hence it sufficesto provethat the Diophantine functionsare closed undercomposition, primitive recursion and minimalization. Composition:If h(xl,**, Xn) =w(gj(xj9 ** I xn), ** m(x * , X.)), where f,g , 9mg are Diophantine,then so is h since .. ... y = h(xl **, Xn) ->(3t19 ** tm) [tj g(XI,*I , Xn) & & tm gm(Xlx * Xn) & y =f(tl, ... tm)] PrimitiveRecursion: If h(xls *-*9Xn,1) = f(XI, -qssXn) ... h(xl **,Xn, t + 1) = g(t,h(xl, ** X*, n t)9 xl ,Xn), andf, g are Diophantine,then (using the sequence number theorem to "code" the numbersh(xl, ... ,xn,1), h(xl Xn,2),*,h(xj,***,x z): y = h (x, I 9.., Xn9 Z) < (3u) {(3v) [v) = S(, u) & v =f(x1,I, xn)] (Vt) z [(t=z) V (3v) (v = S(t + 1,u) & v = g(t, S(t,u), xI, ..,x))] & y = S(z, u)} 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 260 MARTIN DAVIS [March so that(using Theorem 5.1) h is Diophantine. Minimalization:If h(x1, --x mnU[(x1, j..,y) = g(xl, ...,x" y)], wheref,g are Diophantine,then so is h since, y = h(x-,X*> = (3Iz)[z =A(XI, ---Xn,Y)&Z XXI, **?,Xn-Y)] ? (Vt):5y[(t = Y) V (3u,v) (u f(x,, ... 9Xn9t) &v=g(xl,..X ot)&( < v Vv< u)]. 7. A universalDiophantine set. An explicitenumeration of all theDiophantine setsof positive integers will now be described.Any polynomial with positive integer coefficientscan be builtup from 1 andvariables by successi've additions and multiplica- tions.We fixthe alphabet Xo,X1, X2, X3,' of variablesand thenset up the followingenumeration of all such polynomials (usingthe pairing functions): P1I = P3i1 = Xi- P3 - PL(i) + PR(i) P3i+ I PL(i) PR(i) WritePi = Pi(x0, x, * x), wheren is largeenough so thatall variablesoccurring in Pi are included.(Of coursePi will not in generaldepend on all of thesevariables.) Finally,let n= {xoj (3x1 , ...,x") [PL(fn)(X0o * I, Xn) PR(n)(x0 lX 1 .,xX)}. Here, PL(,)and PR(n) do notactually involve all ofthe variables xo, x1,I xn-but clearlycannot involve any others. (Recall that L(n), R(n) ? n.) Bythe way the se- quencePi has beenconstructed, it is seenthat the sequence of sets: DI1, 23,D39 D4, * includesall Diophantinesets. Moreover: THEOREM7.1 (UniversalityTheorem5). { 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] HILBERTS TENTHPROBLEM IS UNSOLVABLE 261 Proof.Once again usingthe sequence number theorem, it is claimedthat: x e Dn. (3u) {S(1, u) = 1 & S(2, u) = x & (Vi):5[S(3i, u) = S(L(i), u) + S(R(i), u)] &(Vi)sn[S(3i + 1,u) = S(L(i), u) S(R(i), u)] & S(L(n), u) = S(R(n), u)}. It is clearenough that the predicate on theright-hand side of thisequivalence is Diophantine,so it is onlynecessary to verifythe claim: Let x E Dn for given x, n. Then there are numberst1, ** *, tn such that PL(n) (X, tl, ***, tn) = QL(n) (X, tl, **., ta). Chooseu (bythe sequence number theorem) so that (*) ~~~~SOS,u) =Pj(x1, t,I *. , tn)3 j = ,2 3n + 2. Then in particularS(2, u) = x and S(3i - 1,u) = ti-1, i = 2,3, *,n + 1. Thus the right-handside of theequivalence is true. Conversely,let theright-hand side holdfor given n, x. Set t, = S(5, u), t2 = S(8, u), ... , t = S(3n + 2, u). Then, (*) mustbe true. Since S(L(n), u) = S(R(n), u), it mustbe the case that = . PL(n)(X4 t3, *s*,tn) pR(n)(X3, t I tn)9 so thatx E Dn* SinceDI, D2,D3, .*', givesan enumerationof all Diophantinesets, it is easyto constructa set different from all ofthem and hence non-Diophantine. That is, define: v = {n fn Dnl} THEOREM7.2. V is notDiophantine. Proof. This is a simple application of Cantor's diagonal method. If V were Diophantine,then for some fixedi, V = Di. Does i E V? We have: i c-V.>i c-Di; i e-V i 0 Di. This is a contradiction. THEOREM 7.3. Thefunctiong(n,x) definedby: g(n,x)=1 if xq Dn, g(n,x)=2 if xeDn, is not recursive. Proof.If g wererecursive then it wouldbe Diophantine(Theorem 6.1), say: 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 262 MARTIN DAVIS [March y = g(n,x) *(3y1,* *, Ym)[P(n, xy,ylY,Y,**,Ym) =0?] Butthen, it wouldfollow that ... V-={x I (3y,,* * , Ym)[P(x, x, 11Yll, Ym)= ?]} whichcontradicts Theorem 7.2. UsingTheorem 7.1, write: Xc- Dn* (3zl **"' Zk) [P(n,x, z, Zk) = O]. whereP is somedefinite (though complicated) polynomial. Suppose there were an algorithmfor testingDiophantine equations for possessionof positiveinteger solutions; i.e., an algorithmfor Hilbert's tenthproblem! Then forgiven n, x this algorithmcould be usedto test whether or not the equation P(n, x, ZI, - z Zk) = has a solution,i.e., whetheror not x E Dn. Thus the algorithmcould be used to computethe function g(n, x). Sincethe recursive functions are just those for which a computingalgorithm exists, g wouldhave to be recursive.This would contradict Theorem7.3, and thiscontradiction proves: THEOREM 7.4. Hilbert's tenthproblem is unsolvable! Naturallythis result gives no informationabout the existenceof solutionsfor any specificDiophantine equation; it merelyguarantees that there is no single algorithmfor testing the class of all Diophantineequations. Also note that: E- **' = X v - (3z,, Zk) [P(X, X, ZI, ", ZJ) 0] ... J(3z ", ,ZO [P(X, X, Z1s Zk) = 01 -+ I = 0} (VZ1,** Zk) [P(X, X, Z 1, 9)Zk > ? V P(x, X, Z, ..., Zk) < 0] whichshows that if either - or unboundeduniversal quantifiers (Vz) or implication ( ) are permittedin the language of Diophantine predicates, then non-Diophantine sets will be produced. It is naturalto associatewith each Diophantineset a dimensionand a degree; i.e.,the dimension of S is theleast n forwhich a polynomialP existsfor which: (*) S-{xl (3y1, Yn) [P(X,Y Yn = ?]} and thedegree of S is theleast degree of a polynomialP satisfying(*) (permittingn to be as largeas onelikes). Now it is easyto see: THEOREM 7.5. EveryDiophantine set has degree ? 4. Proof The degreeof P satisfying(*) maybe reducedby introducing additional 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 263 variablesZj satisfyingequations of the forim Zi YiYk zj zi= Yi Zj = XYi zj = x By successivesubstitutions of thezj's intoP itsdegree can be broughtdown to 2. Hence the equationis equivalentto a systemof simultaneousequations each of degree2. Summingthe squares gives an equationof degree 4. A lesstrivial (and more surprising) fact is: THEOREM 7.6. There is an integer m such that every Diophantine set has dimension< m. Proof. Write Dn= {xl (3YI,.. Ym)[P(x n,yl, .. IYm) = 0]}s whichis possibleby the universality theorem. Then the dimensionof Dn is .. Sq = {X I (3y1 , yq) [X = (Yt + 1) (Yq + 1)]}. HereS2 isthe set of composite numbers; Sq is theset of "q-fold" composite numbers. It is surelysurprising that it is possibleto givea Diophantinedefinition of Sq (for large q) requiringfewer than q parameters(cf. [19]). How largeis m, thenumber of parametersin the universalDiophantine set? A directcalculation using the arguments given here would yield a numberaround 50. ActuallyMatiyacevic and JuliaRobinson have veryrecently shown that m = 14 will suffice! Theunsolvability ofHilbert's tenth problem can be used to obtaina strengthened formof Godel's famousincompleteness theorem: THEOREM 7.7. Correspondingto any given axiomatizationof numbertheory, thereis a Diophantineequation whichhas no positiveinteger solutions, but such that thisfact cannot be provedwithin the given axiomatization. A rigorousproof would involve a precisedefinition of "axiomatization of number theory"which is outsidethe scope of thisarticle. An informalheuristic argument follows: One usesthe given axiomatization to systematicallygenerate all ofthe theorems (i.e., consequencesof the axioms).Among these theorems will be someasserting 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 264 MARTIN DAVIS [March thatsome Diophantine equation has no solution.Whenever such is encounteredit is placedon a speciallist called LISTA. At the same time a list,LIST B, is made of Diophantineequations which have solutions.LIST B is constructedby a search procedure,e.g., at thenth stage of the search look at the first n Diophantineequations (in a suitablelist) and test for solutions in whicheach argument is < n. Thus every Diophantineequation which has positive integer solutions will eventually be placedin LIST B. If likewiseeach Diophantineequation with no solutionswould eventually appearin LIST A, thenone wouldhave an algorithmfor Hilbert's tenth problem. Namely,to test a givenequation for possession of a solutionsimply begin generating LIST A and LIST B untilthe given equation appears in onelist or the other.Since Hilbert'stenth problem is unsolvable,some equationwith no solutionmust be omittedfrom LIST A. Butthis is justthe assertion of thetheorem. 8. Recursivelyenumerable sets. It is nowtime to settlethe questionraised at the beginning:which sets are Diophantine? DEFINITION. 8.1. A set S of n-tuplesof positiveintegers is called recursively enumnerableif thereare recursivefunctions f(x,x1, 1 ., x), g(x,gx1. -qXn) suchthat: S = { 4 (3u)[P(x1, ... x , SO1 u), u..,S(M,U)) = Q(X* ,* xu), **S(m, SO )] so thatS is recursivelyenumerable. Converselyif S is recursivelyenumerable there are recursivefunctions f(x,x1, .. ,x"), g(x,xI,--- ,xn) suchthat . .* (3x,z) [z =Xf(x,x1, .*X,xn) &z = (X,x,,x")].n Thus by Theorem6.1, S is Diophantine. 9. Historicalappendix. The presentexposition has ignoredthe chronological orderin whichthe ideas weredeveloped. The firstcontribution was by Godel in his celebrated1931 paper [16]. The mainpoint of Godel's investigationwas the existenceof undecidablestatements in formalsystems. The undecidablestatements Godel obtainedinvolved recursive functions, and in orderto exhibitthe simple number-theoreticcharacter of thesestatements, Godel usedthe Chinese remainder theoremto reducethem to "arithmetic"form. The techniqueused is just whatis 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] HI'LBERT'S TENTH PROB3LEMIS UNSOLVABLE 265 usedhere in provingTheorem 1.3 (the sequence number theorem) and Theorem6.1 (in the direction:every recursive function is Diophantine).However without the techniquesfor dealing with bounded universal quantifiers as discussedin thispaper, the bestresult yielded by Godel's methodsis thatevery recursive function (and indeedevery recursively enumerable set) can be definedby a Diophantineequation precededby a finitenumber of existential and bounded universal quantifiers6. In my doctoraldissertation (cf. [5], [6]), I showedthat all butone of the bounded universal quantifierscould be eliminated,so thatevery recursively enumerable set S could be definedas S = {x I (3y) (Vk) D = { .. .. - (3u,, ..X Ung VI, * * Vng Wl, * * Wn) EP(X 1 **. *,XWsU19 ' Un1V19-,** 1Vni WD -9%*W) = O &u1 =vI '&. &u" = Vln]. In particular,the functions(k) and n! were shown by her to be exponential 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 266 MARTIN DAVIS [March Diophantine.This is reallywhat is shownin proving(1) and (2) of Theorem4.1. The presentproof of (2) is justhers; the proof of (1) givenhere is a simplifiedvar- iantof thatin [27]. (It is due independentlyto Julia Robinson and Matiyasevic.) Theidea ofusing the Chinese remainder theorem to codethe effect of a bounded universalquantifier first occurred in thework of myself and Putnam[7]. In [8], we refinedour methods and were able to show,beginning with the Davis normalform, thatIF thereare arbitrarilylong arithmeticprogressions consisting entirely of primes(still an openquestion), then every recursively enumerable set is exponential Diophantine.In our proofwe needed to establishthat h(a, b,y) = fly= (a + bk) is exponentialDiophantine, which we did extendingJulia Robinson's methods. (The proofgiven here of (3) of Theorem4.1 is a muchsimplified argument found much laterby Julia Robinson-cf. [29].) JuliaRobinson then showed first how to eliminate the hypothesisabout primesin arithmeticprogression, and thenhow to greatly simplifythe proof along the lines of Lemma5.2 ofthis article. Thus we obtainedthe theoremof [9] thatevery recursively enumerable set is exponentialDiophantine. Attentionwas nowfocused on theJulia Robinson hypothesis since it was plain thatit wouldimply that Hilbert's tenth problem was unsolvable. Manyintelesting propositions were found to implythe Julia Robinson hypoth- esis.7. Howeverthe hypothesisseemed implausible to many,especially because it wasrealized that an immediateand surprising consequence would be theexistence of an absoluteupper bound for the dimensions of Diophantinesets (cf. Theorem 7.6). Thus in his review[19] Kreisel said concerningthe resultsof [9]: "... it is likelythe presentresult is notclosely connected with Hilbert's tenth problem. Also it is not altogetherplausible that all (ordinary)Diophantine problems are uniformlyreducible to thosein a fixednumber of variables of fixed degree.... " The JuliaRobinson hypothesis was finallyproved by Matiyasevic[23], [24]. Specificallyhe showedthat if we define d-=a2 = 1, an+1 = an + an-1 so thatan is thenth Fibonacci number, then the function a2n is diophantine.Then since,for n > 3, as is easilyseen by induction, )< an <2n2 the set D = { 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 PROBLEMIS UNSOLVABLE 267 In particular,itwas Matiyasevicwho taught us howto use resultslike Lemmas2.11, 2.12,and 2.22of the present exposition. (Matiyasevic himself used analogous results forthe Fibonacci numbers.) It was soonnoticed (by S. Kochen)that by a simpleinductive argument the use ofthe Davis normalform could now be entirelyavoided, as has been done in the presentexposition. Let #(P) be thenumber of solutionsof theDiophantine equation P = 0. Thus O? #(P) ? 8R. Hilbert'stenth problem seeks an algorithmfor deciding of a given P whetheror not #(P) = 0. But there are many related questions: Is there an algorithmfor testing whether #(P) = N, or #(P) = 1, or #(P) is even? I was able to showeasily (beginning with the unsolvability of Hilbert's tenth problem) that all ofthese problems are unsolvable. In factif A = {0,1,2,3, 4o} and B ' A, B # 0, B # A,then one can readilyshow that there is no algorithmfor determiningwhether or not#(P) E B (cf.[15]). Thefact that no generalalgorithm such as Hilbertdemanded will be forthcoming adds to the interestof algorithms for dealing with special classes of Diophantine equations.Alan Bakerand his coworkers[1], [2] havein recentyears made con- siderableprogress in thisdirection. Notes 1. Thesepairing functions (but of coursenot theirbaing Diophantine) were used by Cantorin his proofof thecountability of therational numbers. J. Robertsand D. Siefkeseach correctedan errorin thedefinition of these functions. They, as wellas W. Emerson,M. Hausner,Y. Matiyasevic, and JuliaRobinson made helpfulsuggestions. 2. For example,cf. [25], pp. 175-180.Matiyasevic used instead the equations x2 - xy- = 1, U2- muv + V2 1. 3. The recursivefunctions are usuallydefined on thenonnegative integers. This createsa minor butannoying technical problem in comparingthe present definition with one in theliterature (e.g., cf. [6],p. 41; also Theorem4.2 on p. 51). Thus one can simplynote that f(xl,..., x") is recursivein the presentsense if and onlyif f(t1 + 1,. *,tn + 1) - 1 is recursivein the usual sense.From the pointof viewof theintuitive "computability" of thefunctions involved this doesn't matter at all; oneis simplyin theposition of using the positive integers as a "code" forthe nonnegative integers - using n + 1 to representn. 4. Inclusionof S (i, u) in thislist is redundant.That is, S (i, u) can bs obtainedusing our three operationsfrom the remaininginitial functions. 5. The methodof proof is JuliaRobinson's, [28], [30]. If onewere permitted to use theenumera- tiontheorem in recursivefunction theory ([6], p. 67. Theorem1.4), the Universality Theorem would followat once fromTheorem 6.1. 6. Actuallythe result which G6del stated(as opposedto whatcan be obtainedat once byuse of his techniques)was somewhatweaker. Indeed, the very definition of theclass of recursivefunctions and theperception of theirsignificance came severalyears later in thework of G6del,Church, and Turing.In particularths sugg-stionthat recursiveness was a precise equivalentof the intuitive 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 268 MARTIN DAVIS [March notionof beingcomputable by an explicitalgorithm was made independentlyby Churchand by Turing.And of courseit is thisidentification which is essentialin regardingthe technicalresults discussedin thisaccount as constitutinga negative solution of Hilbert's tenth problem. (For further discussionand references,cf. [6].) 7. For example,I showed([13]) thatthe Julia Robinson hypothesis would follow from the non- existenceof nontrivialsolutions of theequation 9 (U2 -!- 7i2)2 __ 7(x2 + 7y2)2 - 2. The methodsused readilyshow that the same conclusionfollows if theequation has onlyfinitely manysolutions. Cudnovskii [4] claimsto haveproved that 2x is diophantine(and hence the Julia Robinsonhypothesis) using this equation. Apparently there is a possibilitythat some of Cudnovskii's workmay have been done independently ofMatiyasevic - butI havenot been able to obtaindefinite informationabout this. References 1. Alan Baker,Contributions to thetheory of Diophantine equations: I. On therepresentation of integersby binaryforms, II. The Diophantineequation y2 xX3 + k, Philos.Trans. Roy. Soc. London Ser. A, 263 (1968) 173-208. 2. Alan Baker,The Diophantineequation y2 == ax3 F bx2 + cx + d, J.London Math. Soc., 43 (1968) 1-9. 3. G. V. Cudnovskii,Diophantine predicates (Russian), Uspehi Mat. Nauk, 25 (1970) no. 4 (154), pp. 185-186. 4. , Certainarithmetic problems (Russian), Ordena Lenina Akad. Ukrains. SSR, Preprint IM-71-3. 5. MartinDavis, Arithmeticalproblems and recursivelyenumerable predicates, J. Symbolic Logic, 18 (1953) 33-41. 6. , Computabilityand Unsolvability,McGraw Hill, New York, 1958. 7. MartinDavis and HilaryPutnam, Reduction of Hilbert'stenth problem, J. SymbolicLogic, 23 (1958) 183-187. 8. -and -, On Hilbert'stenth problem, U. S. AirForce 0. S. R. ReportAFOSR TR 59-124(1959), Part III. 9. MartinDavis, HilaryPutnam, and JuliaRobinson, The decisionproblem for exponential Diophantineequations, Ann. Math.,74 (1961) 425-436. 10. MartinDavis, Applicationsof recursivefunction theory to numbertheory, -Proc. Symp. Pure Math.,5 (1962) 135-138. 11. , Extensionsand corollariesof recentwork on Hilbert'stenth problem, Illinois J. Math.,7 (1963) 246-250. 12. MartinDavis and HilaryPutnam, Diophantine sets over polynomial rings, Illinois J. Math., 7 (1963)251-256. 13. MartinDavis, One equationto rulethem all, Trans.New York Acad. Sci., SeriesII, 30 (1968) 766-773. 14. , An explicitDiophantine definition of theexponential function, Comm. Pure Appl. Math.24 (1971) 137-145. 15. , On the numberof solutionsof Diophantineequations, Proc. Amer.Math. Soc., 35 (1972) 552-554. 16. KurtGodel, Uber formal unentscheidbare Satze der Principia Mathematica und verwandter SystemeI, Monatsh.Math. und Physik,38 (1931) 173-198.English translations: (1) Kurt,Godel, On FormallyUndecidable Propositions of PrincipiaMathematica and Related Systems,Basic 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] HILBERTS TENTH PROBLEMIS UNSOLVABLE 269 Books, 1962. (2) MartinDavis (editor),The Undecidable,Raven Press,1965, pp. 5-38. (3) Jean Van Heijenoort(editor), From Frege to Gddel,Harvard University Press, 1967, pp. 596-616. 17. G. H. Hardyand E. M. Wright,An Introductionto theTheory of Numbers, Fourth edition, OxfordUniversity Press, 1960. 18. David Hilbert,MathematichseProbleme, Vortrag, gehalten auf dem internationalen Mathe- matiker-Kongresszu Paris 1900. NachrichtenAkad. Wiss. Gottingen,Math. -Phys.Kl. (1900) 253-297.English translation: Bull. Amer.Math. Soc., 8 (1901-1902)437-479. 18a. N. K. Kosovskii,On Diophantinerepresentations ofthe solutions of Pell's equation(Rus- sian), Zap. Nau6n. Sem. Leningrad.Otdel. Mat. Inst. Steklova,20 (1971) 49-59. 19. GeorgKreisel, Review of [9]. Mathematical Reviews, 24 (1962)Part A, p. 573(review number A 3061). 20. YuriMatiyasevi6, The relationof systems of equations in wordsand theirlengths to Hilbert's tenthproblem (Russian). Issledovaniya po KonstruktivnoiMatematike i MatematiceskoiLogike II. Vol. 8, pp. 132-144. 21. - , Two reductionsof Hilbert'stenth problem (Russian), Ibid., pp. 145-158. 22. , Arithmeticrepresentation ofexponentiation (Russian). Ibid., pp. 159-165. 23. , Enumerablesets are Diophantine(Russian), Dokl. Akad. Nauk SSSR, 191 (1970) 279-282.Improved English translation: Soviet Math. Doklady, 11 (1970) 354-357. 23a. , Diophantinerepresentation of theset of primenumbers (Russian). Dokl. Akad. Nauk SSSR, 196 (1971) 770-773. ImprovedEnglish translation with Addendum: Soviet Math. Doklady,12 (1971) 249-254. 23b. -, Diophantinerepresentation of recursivelyenumerable predicates, Proc. Second ScandinavianLogic Symp.,editor, J. E. Fenstad,North-Holland, Amsterdam, 1971. 23c. -, Diophantinerepresentation of recursivelyenumerable predicates, Proc. 1970 Intern.Congress Math., pp. 234-238. 24. - , Diophantinerepresentation of enumerablepredicates (Russian), Izv. Akad. Nauk SSSR, Ser. Mat. 35 (1971) 3-30. 24a. , Diophantinesets (Russian), Uspehi Mat. Nauk,27(1972) 185-222. 25. Ivan Nivenand HerbertZuckerman, An Introductionto theTheory of Numbers,2nd ed., Wiley,New York, 1966. 26. HilaryPutnam, An unsolvableproblem in numbertheory, J. Symb. Logic, 25 (1960)220-232. 27. JuliaRobinson, Existential definability in arithmetic,Trans. Amer. Math. Soc., 72 (1952) 437-449. 28. ^, Diophantinedecision problems, MMA Studiesin Mathematics,6 (1969) [Studies in NumberTheory, edited by W. J. LeVeque, pp. 76-116]. 29. , UnsolvableDiophantine problems, Proc. Amer.Math. Soc., 22 (1969) 534-538. 30. -, Hilbert'stenth problem, Proc. Symp.Pure Math., 20 (1969) 191-194. 31. RaphaelM. Robinson,Arithmetical representation ofrecursively enumerable sets, J. Symb. Logic, 21 (1956) 162-186. 32. -, Some representationsof Diophantinesets, J. Symb.Logic, forthcoming. 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