David Hilbert (1862-1943) presented a list of 23 mathematical problems at the International Mathe-matical Congress in Paris in 1900. The tenth problem can be stated as follows:
Given a diophantine equation with a finite number of unknowns and with integral coefficients.
Devise a procedure that determines in a finite number of steps whether the equation is solvable in integers.
In 1970, a result in mathematical logic known as Matiyasevich’s theorem settled the problem negatively.
Let Z[X1, X2, . . . , Xn] denote the commutative polynomial ring in the unknowns X1, X2, . . . , Xn
with integer coefficients. Each polynomialpin Z[X1, X2, . . . , Xn] gives rise to adiophantine equation
p(X1, X2, . . . , Xn) = 0 (8.38)
asking forintegersolutions of this equation. By the fundamental theorem of algebra, every non-constant single-variable polynomial with complex coefficients has at least one complex root, or equivalently, the field of complex numbers is algebraically closed.
Example 8.27.Linear diophantine equations have the forma1X1+. . .+anXn=b. Ifbis the greatest common divisor of a1, . . . , an (or a multiple of it), the equation has an infinite number of solutions.
This is Bezout’s theorem and the solutions can be found by applying the extended Euclidean algorithm.
On the other hand, ifb is not a multiple of the greatest common divisor ofa1, . . . , an, the diophantine
equation has no solution. ♦
Letpbe a polynomial inZ[X1, . . . , Xn]. Thenatural variety ofpis the zero set ofpinNn0, V(p) ={(x1, . . . , xn)∈Nn0 |p(x1, . . . , xn) = 0}. (8.39) Each polynomial function is defined by composition of addition and multiplication of integers and so leads to the following result. For this, it is assumed that addition and multiplication of integers are computable functions.
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Proposition 8.28.Each natural variety is a decidable set.
A diophantine set results from a natural variety by existential quantification. More specifically, let pbe a polynomial inZ[X1, . . . , Xn, Y1, . . . , Ym]. A diophantine set is ann-ary relation
{(x1, . . . , xn)∈Nn0 |p(x1, . . . , xn, y1, . . . , ym) = 0 for somey1, . . . , ym∈N0}, (8.40) which will subsequently be denoted by
∃y1. . .∃ym[p(x1, . . . , xn, y1, . . . , ym) = 0]. (8.41) Proposition 7.15 yields the following assertion.
Proposition 8.29.Each diophantine set is semidecidable.
Example 8.30.
• The set of positive integers is diophantine, since it is given by{x| ∃y[x=y+ 1]}.
• The predicates ≤and<are diophantine, sincex≤y if and only if∃z[y=x+z], and x < yif and only if∃z[y=x+z+ 1].
• The predicatea≡bmodcis diophantine, since it can be written as∃x[(a−b)2=c2x2]. ♦ The converse of the above proposition shown by Yuri Matiyasevich (born 1947) in 1970 is also valid, but the proof is not constructive.
Theorem 8.31.Each semidecidable set is diophantine.
That is, for each semidecidable setA in Nn0 there is a polynomialpin Z[X1, . . . , Xn, Y1, . . . , Ym] such that
A=∃y1. . .∃ym[p(x1, . . . , xn, y1, . . . , ym) = 0]. (8.42) The negative solution of Hilbert’s tenth problem can be proved by using the four-square theorem due to Joseph-Louis Lagrange (1736-1813). For this, an identity due to Leonhard Euler (1707-1783) is needed which is proved by multiplying out and checking:
(a2+b2+c2+d2)(t2+u2+v2+w2) = (8.43)
(at+bu+cv+dw)2+ (au−bt+cw−dv)2+ (av−ct−bw+du)2+ (aw−dt+bv−cu)2. This equation implies that the set of numbers which are the sum of four squares is closed under multiplication.
Theorem 8.32 (Lagrange). Each natural number can be written as a sum of four squares.
For instance, 3 = 12+ 12+ 12+ 02, 14 = 32+ 22+ 12+ 02, and 39 = 52+ 32+ 22+ 12.
Proof. By the above remark it is enough to show that all primes are the sum of four squares. Since 2 = 12+ 12+ 02+ 02, the result is true for 2.
Letpbe an odd prime. First, claim that there is some numbermwith 0< m < psuch that mpis a sum of four squares. Indeed, consider thep+ 1 numbersa2 and−1−b2 where 0≤a, b≤(p−1)/2.
Two of these numbers must have the same remainder when divided byp. However,a2 andc2have the
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same remainder when divided by pif and only if pdivides a2−c2; that is, p divides a+c or a−c.
Thus the numbersa2 must all have different remainders. Similarly, the numbers−1−b2have different remainders. It follows that there must bea andb such that a2 and−1−b2 have the same remainder when divided byp; equivalently, a2+b2+ 12+ 02 is divisible byp. Butaandbare at most (p−1)/2 and soa2+b2+ 12+ 02has the formmp, where 0< m < p. This proves the claim.
Second, claim that if mp is the sum of four squares with 1 < m < p, there is a number n with 1≤n < msuch thatnpis also the sum of four squares. Indeed, letmp=x21+x22+x23+x24and suppose first thatmis even. Then either eachxiis even, or they are all odd, or exactly two of them are even. In the last case, it may be assumed thatx1andx2 are even. In all three cases each ofx1±x2andx3±x4
are even. So (m/2)pcan be written as ((x1+x2)/2)2+ ((x1−x2)/2)2+ ((x3+x4)/2)2+ ((x3−x4)/2)2, as required.
Next letmbe odd. Define numbers yi byxi≡yimodmand|yi|< m/2. Theny21+y22+y32+y24≡ x21+x22+x23−x24modm and soy21+y22+y32+y24 =nm for some numbern ≥0. The casen = 0 is impossible, since this would make every yi zero and so would make everyxi divisible bym. But then mpwould be divisible bym2, which is impossible sincepis a prime and 1< m < p.
Clearly,n < msince eachyiis less thanm/2. Note thatm2np= (x21+x22+x23−x24)(y21+y22+y32+y42).
Use Euler’s identity to write m2np as a sum of four squares. Claim that each integer involved in this representation is divisibe bym. Indeed, one of the involved squares is (x1y1+x2y2+x3y3+x4y4)2. But the sumx1y1+x2y2+x3y3+x4y4is congruent modmtoy21+y22+y23+y42, sincexi≡yimodm. However, y12+y22+y32+y24≡0 modm, as needed. Similarly, the other three integers involved are divisible bym.
Now m2np is the sum of four squares each of which is divisible by m2. It follows that np itself is the sum of four squares, as required.
Finally, the process to write a multiplempofpas a sum of four primes iterates leading to smaller
multiplesnp ofp. This process will end by reachingp. ⊓⊔
Letpbe a polynomial inZ[X1, . . . , Xn]. Define the integral polynomialqin the unknownsT1, . . . , Tn, Conversely, if (t,u,v,w)∈Z4n is a solution of the diophantine equation (8.47), then x∈Nn0 defined as in (8.46) provides an integer solution of the diophantine equation (8.45).
It follows that if there is an effective procedure to decide whether the diophantine equation (8.47) hasinteger solutions, there is one to decide if the diophantine equation (8.45) hasnon-negative integer solutions.
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Theorem 8.33.Hilbert’s tenth problem is undecidable.
Proof. The setK ={x|x∈domφx} is recursively enumerable and so by Matiyasevich’s result there is a polynomialpin Z[X, Y1, . . . , Ym] such that
K=∃y1. . .∃ym[p(x, y1, . . . , ym) = 0]. (8.48) Suppose there is an effective procedure to decide whether or not a diophantine equation has non-negative integer solutions. Then the question whether a numberx∈N0 lies inKor not can be decided by finding a non-negative integer solution of the equationp(x, Y1, . . . , Ym) = 0. However, this contradicts
the undecidability ofK. ⊓⊔
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