ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES

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1 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES TERO HARJU AND DIRK NOWOTKA Astract. We prove that an independent system of equations in three variales with a nonperiodic solution and at least two equations consists of alanced equations only. F that, we show that the intersection of two different entire systems contains only alanced equations, where an entire system is the set of all equations solved y a given mphism. Furtherme, we estalish that two equations which have a common nonperiodic solution have the same set of periodic solutions are not independent.. Introduction Systems of wd equations in three variales are investigated in this article. Consider f example the following system S of two equations xyz = zyx and xyyz = zyyx which has a solution α with αx = αz = a and αy =. This solution is called nonperiodic since αx, αy, and αz are not powers of the same wd. The system S is also independent since there exist the solutions β and γ, with βx = a and βy = and βz = aa and γx = a and γy = and γz = aa, which solve either one of the two equations ut not the other. Both equations in this example are alanced, that is, the numer of occurrences of each variale on the left and the right hand side is the same. The main result of this article states that every independent system with at least two equations and a nonperiodic solution, consists of alanced equations only. Let α e a nonperiodic solution. We call the set of all equations solved y α the entire system of equations generated y α. It is shown that the intersection of two different entire systems can only contain alanced equations. Furtherme, we estalish that two equations which have a common nonperiodic solution have the same set of periodic solutions are not independent. These two facts prove the result mentioned aove. Even though this result is interesting in its own right, it also provides me insight into the following question: Does there exist an independent system of three equations in three variales that has a nonperiodic solution? If it does, it must contain alanced equations only. This question was implicitely raised y Culik II and Karhumäki [3] first in 983. This introduction is followed y the preliminaries, Section 2, where the notations f this article are fixed. Section 3 introduces Spehner s [7] characterization of solutions of equations in three variales which is compared with an earlier result Date: June 26, Key wds and phrases. cominatics on wds, systems of equations, independence.

2 2 TERO HARJU AND DIRK NOWOTKA y Budkina and Markov [] in Section 4. The intersection of Spehner s entire systems is investigated in Section 5 which provides the foundation f the proof of the main result in Section 6. This paper ends with concluding remarks in Section Preliminaries In this section we fix the notations f this paper. We refer to [4, 5, 2] f me asic and general definitions. Let X = {x, y, z} e a fixed set of three variales and let X + denote the semigroup of all finite, nonempty wds over X. Let ε denote the empty wd. A wd v is a prefix of u, denoted y v u, if there exists a wd w such that u = vw. If w ε, then v is a proper prefix of u, denoted y v < u. Accdingly, v u and v u denote that v is a suffix and proper suffix of u, respectively, that is u = wv. An equation in X is any pair u, w of wds in X +, usually written as u = w. An equation is called alanced if every variale has the same numer of occurrences on each side. Let e e an equation u = w, then we areviate v u and v 2 w y v, v 2 e, and v u and v 2 w y v, v 2 e. We say that an equation e starts ends with v and v 2, if v, v 2 e v, v 2 e, respectively v 2, v e v 2, v e, respectively. Let A = {a, } e a fixed set of two letters. A solution of an equation u = w is a mphism α: X + A + such that αu = αw. Note, that f any solution in a finite set of letters a solution in A can e found, since any finitely generated semigroup can e emedded in A +. An equation u = w is called reduced if ε is the greatest common prefix and suffix of u and w. Note, that every equation can e transfmed into its reduced fm, that is y dropping common pre- and suffixes, without changing its set of solutions. Two mphisms α: X + A + and β : Y + B + are isomphic if there exist two isomphisms ρ : X + Y + and ρ 2 : A + B + such that ρ 2 α = βρ, that is, the following diagram commutes: X + α A + ρ ρ 2 Y + β B + We call α a permutation of β if X = Y and A = B. A mphism is called periodic if it is isomphic to a mphism in {a} +, that is, there exists a wd w such that αx {w} + f all x X; otherwise it is called nonperiodic. Let v X + and α e as efe, then A α v and B α v denote the numer of occurrences of the letters a and, respectively, in αv. We might areviate αv with v, if the context is clear. A system of equations, system f sht, is a nonempty set of equations. A solution of a system is a mphism that solves all equations in the system. Two systems are equivalent if they have exactly the same set of solutions. A system of equations is called independent if it is not equivalent to any of its proper susets. Let α: X + A + e a mphism. Then the kernel kerα = {u, w αu = αw} = α α of α is also called an entire system generated y α in the context of equations, see [7], denoted y K α. Here, K α consists of all equations f which α is a solution. A suset C of A + is called a ase of C + if C = C + \ C + 2. A suset C of A + is called incontractale if C is a ase of C + and f any D A +, if C + and D + are

3 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 3 isomphic, then v C v v D v. We call α an incontractale mphism if the ase of αx + is an incontractale suset of A +. Consider f example equation xyx = zz which is solved y all of the following mphisms α p defined y p a a p+ a where p 0. Now, α 0, that is and a and a, is incontractale, whereas α p, f any p, is not incontractale. Note, that α p is a principal solution cf. [4] f all p 0. F the rest of this paper we consider nontrivial reduced equations and systems of equations in X and solutions in A + only. 3. A Characterization of Incontractale Solutions In this section we will give Spehner s characterization, see Proposition 2.5 in [7], of nonperiodic incontractale solutions of entire systems. Theem. F every entire system S generated y a nonperiodic mphism, there exists a unique, up to permutation, incontractale nonperiodic mphism α such that S = K α. Let α: X + A + e a nonperiodic incontractale solution, then α is of one of the following types. We define α y a triple αx, αy, αz, and we fix that gcdp +, q + = in the following. Let p, q, then α: a, q+, p+. 2 Let p, q, k and i, j 0 and i + j k, then α: a, a k q, a i a k p a j. 3 Let p > q and i, j < k < i + j, then α: a, a k q, a i a k p q a j. Note that q + p q since gcdp +, q + = y assumption. 4 Let q and i, j k < i + j and n 0 and k t 0, f all t n, then α: a, va k q v, a i a j where v = a k+i+j a k2+i+j a kn+i+j. Spehner gives n in [7], ut this is a misprint. Indeed, we have thet equation xyxyx = zyz implies an entire system with an incontractale solution where and a which is not in any of Spehner s types. 5 Let p, q, i, j and m 0, then α: a, a i+j+m q a j, a i a i+j+m p. In Spehner s paper the second component in αx is a i+j q a j which we think is a typo.

4 4 TERO HARJU AND DIRK NOWOTKA 6 Let q and i, i, j, j 0 and ii = jj = 0 and n 0 and k t 0, f all t n, then α: a, a i a k+i+j a k2+i+j a kn+i+j a j q, a i a j. As f type 4, Spehner gives n in [7], ut this is a misprint. Indeed, the equation x = z implies an entire system with an incontractale solution where y and which would not e in any of Spehner s types. However, this example could elong to case, if p, q 0 there. But then the entire system implied y xy = zx is not contained in any type here since and. So, assuming n 0 in types 4 and 6 closes the gap leaving type as given in [7]. Note, that if i + j < i + j then q = otherwise α does not satisfy any nontrivial equation. 7 In this type α: a,, ε. We oserve that no nonerasing mphism is isomphic to a mphism of this type. Since we consider only nonerasing solutions here, type 7 will not e further investigated. 8 Let p, q, r and i, j 0, then α: a, a i+j+r q a j+r, a i+r a i+j+r p. The following theem completes Spehner s characterization. Theem 2. Any incontractale solution β is equal to a permutation of α in one of the eight previous types. Furtherme, we oserve the following fact. Lemma 3. If α is an incontractale solution then αx = a, up to renaming of a, f some x X. 4. A Comparison of Spehner s and Budkina & Markov s Characterization Let a semigroup that is isomphic to a susemigroup with k generats of a free semigroup e called F -semigroup with k generats. All F -semigroups with 3 generats have een completely characterized y Budkina and Markov in 973 [] and y Spehner in 976 [6]. We use Spehner s presentation on the Semigroups conference in 986 [7] to ase our result on. Unftunately, we had to crect some details of Proposition 2.5 in [7] as mentioned in the previous section. In der to justify these crections we will compare our version of Spehner s Proposition 2.5 with Budkina and Markov s Theem in []. Budkina and Markov estalish that any F -semigroup S with three generats is of one of the following types. i S is isomphic to a susemigroup of {a} +. ii S is isomphic to the free product of an F -semigroup with two generats and an infinite monogenic semigroup, that is, with p, q and gcdp, q =. {a,, c} + {a q,, a p } +

5 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 5 iii S is isomphic to { } + a, a k a kn a j, a i {a, a i a k a kn, a j} + where i, j and i, j, n 0 and k t i and k t j, f all t n, respectively. iv S is isomphic to { a, a k a kn, a i a j} + where i, j and n 0 and k t i + j, f all t n. v S is isomphic to {a, a i a k q a j, a i a k p a j } + where k, p, q and 0 i, i, j, j r and gcdp +, q + = and ii = jj = 0. vi S is isomphic to {a, a k q, a i a j} + where q and i, j k and k < i + j. vii S is isomphic to {a, a k a kn a k q a k a kn, a i a j} + where q and i, j k and k < i + j and n 0 and k t i + j, f all t n. We oserve the following crespondence etween Spehner s and Budkina & Markov s types. Budkina & Markov Spehner ii and 6, where i + j = 0 iii 6 where i + j > 0 and ij = 0 iv 6 where ij > 0 v 2, 3, where p > q +, 5, 8 vi 3, where p = q +, 4, where n = 0 vii 4, where n > 0 Note, that if α is of type 6 and i = j = 0, then αx + is isomphic to {a, } +, note also, that {a} + and {a,, c} + are not of rank Aout the Intersection of Entire Systems In this section we show that entire systems with nonperiodic solutions intersect with alanced equations only. We investigate Spehner s types of entire systems f that purpose. Recall that we require solutions to e nonerasing. So, type 7 will not e considered here. F the rest of this section, let α, β : X + A + e two nonperiodic incontractale mphisms which are not identical up to renaming of a and. Let us fix an equation u = w, denoted y e, and let α and β oth satisfy e. Assume that e is not alanced. We can assume that αx = a without restriction of generality. Let τ α and τ β denote the type of α and β, respectively. In general, let us suscript variales with α and β to indicate the solution they elong to.

6 6 TERO HARJU AND DIRK NOWOTKA The numers of occurences of letters a and will e frequently made use of in the proofs of this section. Therefe, oserve that and which implies u x + A α y u y + A α z u z = w x + A α y w y + A α z w z B α y u y + B α z u z = B α y w y + B α z w z A α zb α y A α yb α z B α y Case: Assume αx = βx = a. Then we have which implies = w x u x u z w z. B β y u y + B β z u z = B β y w y + B β z w z B α y B α z = w z u z u y w y = B βy B β z. Case: Assume αx = βy = a. Then we have B β x u x + B β z u z = B β x w x + B β z w z which implies and 2 B β z B β x = w x u x u z w z A α z A α y B αz B α y = B βz B β x and 3 A β z A β x B βz B β x = B αz B α y. Case: Assume αx = βz = a. Then we otain in the same way 4 A α y A α z B αy B α z = B βy B β x and 5 A β y A β x B βy B β x = B αy B α z. The first lemma states that we can assume α and β to e of a certain shape. Lemma 4. If all equations v = v 2 in K α K β such that αx αx 2 βx 2 βx where x v and x 2 v 2 and x, x 2 X are alanced, then all equations in K α K β are alanced.

7 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 7 Proof. Assume v = v 2 is any equation in K α K β. Case: αx = αx 2 βx = βx 2. Note that if αx = αx 2 and βx = βx 2, then v = v 2 is alanced y assumption. Let αx = αx 2 and βx < βx 2, the other case is symmetric. Assume that v = v 2 is not alanced. Since α and β are incontractale, we have αx = a and αy = αz = up to permutation of letters and variales as can e easily seen from the types of incontractale solutions in Section 3. Now, v x = v 2 x, and we have also βx = a, otherwise the shape of β implies that either v y = v 2 y v z = v 2 z, since B β v = B β v 2, and v = v 2 is alanced, which is a contradiction. But now, equation gives B α y B α z = B βy B β z = ; a contradiction again since we have necessarily gcdb β y, B β z = and also max{b β y, B β z} >, otherwise β is a permutation of α. Case: αx < αx 2 and βx < βx 2. Let α : X + A + and β : X + A + such that α x 2 = αx αx 2 and β x 2 = βx βx 2 and α x 3 = αx 3 and β x 3 = βx 3 f all x 3 X such that x 3 x 2, and let σ : X + X + such that σx 2 = x x 2 and σx 3 = x 3 f all x 3 X such that x 3 x 2. Let v = v 2 e the reduced equation σv = σv 2. Now, α and β are oth solutions f v = v 2 and v = v 2 K α K β and it is easy to see y the shape of σ that v = v 2 is alanced, if, and only if, v = v 2 is alanced. Let α and β e incontractale mphisms such that K α = K α and also K β = K β. Then α is of smaller size than α and β is of smaller size than β. Now, either v = v 2 is alanced we can repeat this construction. In the fmer case we have that v = v 2 is alanced. By repetition of the aove construction, we get a sequence of equations where the size of the cresponding incontractale solutions gets strictly smaller. Therefe, the sequence of equations must end with a alanced equation which implies that also v = v 2 is alanced. Remark 5. Since, y Lemma 4, only equations v = v 2 in K α K β such that αx αx 2 βx 2 βx where x v and x 2 v 2 and x, x 2 X need to e shown to e alanced, the case where αx = βx = a and u = w egins ends with x does not need to e investigated in the following. In the next two susections we investigate the intersection of entire systems of different type first and of the same type then.

8 8 TERO HARJU AND DIRK NOWOTKA 5.. Entire Systems of Different Type. The entire systems K α and K β are assumed to e of different type throughout this susection. Lemma 6. If τ α = τ β = then K α K β contains only alanced equations. Proof. If τ α = then u x = w x which implies βx = a, f, otherwise u y = w y u z = w z, y counting the numer of occurences of in βy and βz, and u = w is alanced. Now, τ β {2, 6} since u = w starts and ends with y and z. But, α implies yx, zx 2 e where x, x 2 {y, z} which implies that τ β 2 since k β. But, also τ β 6 since equation gives q α + p α + = B βy a contradiction since gcdp α +, q α + =. q α + p α + = B β z ; We will not consider type f the rest of this section anyme. Lemma 7. If y, z e and αx = βx = a then K α K β contains only alanced equations. Proof. Since αx = a, we have that αy and αz and α must e of type 2 6, and since βx = a we have βy and βz and β must e of type 2 6. Equation gives that B α y/b α z is an integer if α is of type 6, and if α is of type 2 then B α y/b α z is not an integer since gcdp α +, q α + =. The same holds f B β y/b β z. So, α and β must e of the same type; Lemma 8. If x, y e and αx = βy = a then K α K β contains only alanced equations. Proof. We have a αy and a βx and αz and βz. It follows that τ α = 6 and α is of the shape i α a k,α+j α a kn,α+jα a j α jα where q α, i α and j α, j α 0 and j α j α = 0 and k t,α 0, f all t n α and n α 0, and 6 x i α z, y e. Case: Assume τ β = 2. Then β is of the shape i β a k pβ β a z a k β where p β, q β, k β, i β and 0 and i β + k β. We have that solution β implies xy k β z, y e and we have k β =, y equation 6, a contradiction, xy k β i β x, y e and k β = i β + and equation 3 gives q α n α + = k β = and k β = and = 0. Now, β implies x, yz e and α implies x, xz e x, zz e since j α = 0; qα

9 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 9 Case: Assume τ β = 3. Then β is of the shape i β a k pβ β q β a z a k β where p β > q β and i β, < k β < i β +. Solution β implies xy, y e which contradicts α, see equation 6. Case: Assume τ β = 4. Then β is of the shape i β a k,β+i β + a k n β,β+i β + a k qβ β a k,β +i β + a k n β,β+i β + where q β and i β, < k β < i β + and k t,β 0, f all t n β and n β 0. Now, x i α z, y i β z e and y the shape of βz necessarily i α n β +. If i α = n β + then k β =, and if i α > n β + then k β = i β + ; a contradiction in oth cases. Case: Assume τ β = 5 τ β = 8. Then a βx and βx and a = βy and βz and a βz and e ends in y and z which implies j α = j α = 0 and equation 2 gives B β z B β x = A αy B α y ; Lemma 9. If x, z e and αx = βy = a then K α K β contains only alanced equations. Proof. We have a αz and αy and βx and βz, and it follows that τ β {2, 6}. Case: Assume τ β = 2. Then β is of the shape 7 x a k β z a k β pβ a 8 x a k β pβ a z a k β where 0 k β. Sucase: Assume = 0. Then oth shapes of τ β are symmetric and we consider only case 7. Note, equations solved y β end with x and z which gives τ α {3, 4, 6}. If τ α = 3 then k α = j α since β implies x, zy e ut this contradicts the definition of type 3 where k α > j α. If τ α = 4 τ α = 6 then equation 3 gives B α y = q β + k β p β q β ;

10 0 TERO HARJU AND DIRK NOWOTKA a contradiction since gcdp β +, q β + =. Sucase: Assume > 0 and β is of shape 7. Then e ends with y and z, and τ α = 6. Solution β implies now xy, zy k β e and k β = αy = y the shape of α. If k β = then equation 3 gives B α y = q β + k β p β + ; a contradiction since gcdp β +, q β +. If αy = then equation 2 gives q β + p β + = i α ; a contradiction since gcdp β +, q β + =. Sucase: Assume > 0 and β is of shape 8. Then e ends with x and y, and τ α {5, 6, 8}. If τ α = 5 τ α = 8 then α implies x, zx e and β implies x, zy e; If τ α = 6 then β implies xy k β, zy e and k β = y the shape of α. Now, equation 3 gives k β = q α n α + ; Case: Assume τ β = 6. Then β is of the shape 9 0 k,β+ a k n β,β+ a j β x a k,β+ a k qβ n β,β+ a j β where q β, i β and, j β 0 and j β = 0 and k t,β 0, f all t n β and n β 0. Sucase: Assume β is of shape 9. Then equation 3 gives B α z B α y = A βz A β xb β z which implies B α y =, ut then τ α = 6; Sucase: Assume β is of shape 0. Then equation 3 gives B α z B α y = A βx B β x

11 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES where and j β = 0 and e ends with y and z which implies τ α = 2 with i α and j α = 0. So, α is of the shape y a kα q α iα a kα p α where i α k α. If i α = k α then equation 2 gives q β + = k p α + α q α + a contradiction since gcdp α +, q α + =. If i α < k α then α implies that x kα y, x kα iα z e which gives q β = = and n β = 0 But now, equation 3 gives p α + q α + = ; a contradiction since gcdp α +, q α + =. Lemma 0. If y, z e and αx = βy = a then K α K β contains only alanced equations. Proof. Since αx = a we have αy and αz and α must e of type 2 6, and since βy = a we have a βz and βx. Case: Assume τ α = 2. Then i α = 0 and α is of the shape y a kα q α z a kα p α a j α where p α, q α, k α and j α 0 and j α k α. Now, α implies yx kα, zx kα jα e and 2 x kα jα yx jα, x kα jα z e. Sucase: Assume τ α = 2 and τ β = 3. Then β is of the fm x a k β i β a k pβ β q β a where i β, < k β < i β + and equations solved y β end with y and z, which implies j α = 0. But now, y and 2, we have i β = = and < k β < 2; a contradiction. Sucase: Assume τ α = 2 and τ β = 4. Then β is of the fm k,β+i β + a k n β,β+i β + a k qβ β a k,β +i β + a k n β,β+i β + i β a

12 2 TERO HARJU AND DIRK NOWOTKA where q β and i β, k β < i β + and n β 0 and equations solved y β end with y and z, which implies j α = 0. From and 2 follows that i β = = k β = and n β = 0 and also k α =. Equation 2 gives B β x = q α + p α q α ; a contradiction since gcdq α +, p α + =. Sucase: Assume τ α = 2 and τ β = 5 τ β = 8. Then an equation solved y β ends in x and y, and hence, cannot e solved y α. Sucase: Assume τ α = 2 and τ β = 6 and β is of the fm k,β+i β + a k qβ n β,β+i β + a j β i β a where i β and, j β 0 and q β and j β = 0. Note, that actually j β = 0 since equations solved y α end with y and z x and z. If k α = j α then equation 2 gives B β x = q α + k α p α + ; a contradiction since gcdq α +, p α + =. Hence, k α > j α and α implies yx, zx e, ut f equations solved y β we have y, zz e y, zy e; Sucase: Assume τ α = 2 and τ β = 6 and β is of the fm i β a k,β + a k n β,β+ a j β where i and, j β 0 and q β and j β = 0. Note, that actually = 0 since equations solved y α end with y and z x and z. Equation 3 gives p α + q α + = A βz ; a contradiction since gcdp α +, q α + =. Case: Assume τ α = 6. Then i α = i α = 0 and α is of the shape k,α+jα a knα,α+jα a j α jα where q α and j α, j α 0 and j α j α = 0 and k t,α 0, f all t n α and n α 0. We oserve f later considerations that if 3 y s x, z t y e with s, t, and if q α > s > then we have a jα nα a j α αu and a j α nα+ αw and jα = j α = 0 and k t,α = 0, f all t n α. But, then u x = w x and since βy = a we have also u z = w z and e is alanced; Hence, q α = s = and then a jα nα a j α a αu and qα

13 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 3 a j α nα+ αw and jα > 0 and j α = 0 and equations solved y α end with x and z. Sucase: Assume τ α = 6 and τ β = 2. Then β is of the fm x a k β i β a k pβ β a where p β, q β, k β, i β and 0 and i β + k β. Solution β implies that y i β x, zy k β i β z e y i β x, zy k β x e. If k β = i β + then equation 3 gives B α y = q β + k β p β + ; a contradiction since gcdp β +, q β + =. Therefe, y i β x, zy e which implies that e ends with x and z y equation 3, and hence, = 0 and i β =, since s = f equation 3. Now, either yx, xz e yx, yz e. In the previous case equation 3 implies B α y = q β + p β + ; a contradiction since gcdp β +, q β + =. In the latter case we have then a j nα α a αu and a j nα α a j α αw and n α = 0, since j α > 0, and we have B α y =. Equation 3 gives k β p β q β q β + + = ; Sucase: Assume τ α = 6 and τ β = 3. Then β is of the fm x a k β i β a k pβ β q β a where i β, and k β < i β + and p β > q β. Equation 3 gives B α y = k pβ q β β q β + + i β + and p β < q β since i β + > k β ; Sucase: Assume τ α = 6 and τ β = 4. Then β is of the fm k,β+i β + a k n β,β+i β + a k qβ β a k,β +i β + a k n β,β+i β + i β a where q β and i β, k β < i β + and n β 0. Equations solved y β end with y and z which implies j α = j α = 0. Equation 2 gives A αy B α y = B β x ;

14 4 TERO HARJU AND DIRK NOWOTKA Sucase: Assume τ α = 6 and τ β = 5 τ β = 8. Then β is of the fm x a i β+ +m β q β a i β a i β+ +m β p β x a i β+ +r β q β a +r β i β +r β a i β+ +r β p β where p β, q β, i β,, r β and m β 0. Solution β implies yx, y e which contradicts solution α. Lemma. If y, z e and αx = βz = a then K α K β contains only alanced equations. Proof. We have αy and αz and βx and a βy. It follows that τ α {2, 6}. Case: Assume τ α = 2. Then i α = 0 and α is of the shape y a kα q α z a kα p α a j α where p α, q α, k α and j α 0 and k α j α. Now, α implies 4 yx kα, z e. Sucase: Assume τ α = 2 and τ β = 3. Then k β = since equations solved y β egin with yz k β, z e; a contradiction y the definition of type 3. Sucase: Assume τ α = 2 and τ β = 4. Then β is of the fm k,β+i β + a k n β,β+i β + a k qβ β a k,β +i β + a k n β,β+i β + i β a where p β > q β and i β, k β < i β + and n β 0 and k t,β 0, f all t n β. Equation 4 gives y the shape of β that yx kα, z i β x e which implies either = k iβ,β + i β +, if i β n β, k β = ; a contradiction in oth cases. Sucase: Assume τ α = 2 and τ β = 5 τ β = 8. Then β is of the fm x a i β+ +m β a x a i β+ +r β a +r β i β a i β+ +m β p β i β +r β a i β+ +r β p β where p β, q β, i β,, r β and m β 0, and β implies yz, z e which contradicts 4. Sucase: Assume τ α = 2 and τ β = 6. Then β is of the fm k,β+i β + a k qβ n β,β+i β + a j β 5 i β a

15 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 5 6 i β a k,β + a k n β,β+ a j β where q β and i β, i β,, j β, n β 0 and i β i β = j β = 0 and k t,β 0, f all t n β. If β is of shape 5 then 4 implies yx, z i β x e. We get either = k,β + i β +, a contradiction, n β = = j β = 0 and equation 5 gives q α + p α + = i β ; a contradiction since gcdp α +, q α + =. If β is of shape 6 then equation 5 gives q α + p α + = A βy B β y ; a contradiction since gcdp α +, q α + =. Case: Assume τ α = 6. Then α is of the shape k,α+jα a knα,α+jα a j α jα where j α, j α, n α 0 and j α j α = 0 and k t,α 0, f all t n α. Sucase: Assume τ α = 6 and τ β {2, 3, 5, 8}. Then equation 4 gives q β + p β + = A αy j α B α y ; a contradiction since gcdp β +, q β + =. Sucase: Assume τ α = 6 and τ β = 4. Then β is of the fm k,β+i β + a k n β,β+i β + a k qβ β a k,β +i β + a k n β,β+i β + i β a where q β and i β, k β < i β + and n β 0 and k t,β 0, f all t n β. Equations solved y β end with y and z, hence, j α = j α = 0 and equation 4 gives q β + p β + = A αy ; a contradiction since gcdp β +, q β + =. Lemma 2. If x, y e and αx = βz = a then K α K β contains only alanced equations. Proof. We have a αy and αz and βx and βy. It follows that τ α = 6. Equation 4 gives B β y B β z = A αy A α zb α y qα

16 6 TERO HARJU AND DIRK NOWOTKA which implies τ β = 4 and either a βx a βy since βz = a; Lemma 3. If x, z e and αx = βz = a then K α K β contains only alanced equations. Proof. We have a αz and a βx and αy and βy. Case: Assume τ α = 2. Then α is of the shape y a kα q α iα a kα p α a j α where i α and j α 0 and i α + j α k α. Solution α implies 7 8 x iα y, zx kα iα y e x iα y, zx kα iα jα z e. Sucase: Assume τ α = 2 and τ β = 3. Then β is of the fm i β a k pβ β a y a k β where i β, < k β < i β +. Solution β implies xz k β, z e which contradicts 7 and 8. Sucase: Assume τ α = 2 and τ β = 4. Then β is of the fm i β a k,β+i β + a k n β,β+i β + a k qβ β a k,β +i β + a k n β,β+i β + where q β and i β, k β < i β + and n β 0 and k t,β 0, f all t n β. Solution β implies x, z i β y e and k α = i α and j α = 0. But, equations solved y α end with y and z, and equations solved y β end with x and z; Sucase: Assume τ α = 2 and τ β = 5 τ β = 8. Then β is of the fm i β a i β+ +m β p β i β +r β a i β+ +r β p β y a i β+ +m β a y a i β+ +r β a +r β where p β, q β, i β,, r β and m β 0. Solution β implies zy, z e and j α = 0, and α implies xy, z e; Sucase: Assume τ α = 2 and τ β = 6. Then β is of the fm i β a k,β+i β + a k n β,β+i β + a j β

17 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 7 x y a k,β+ a a k qβ n β,β+ a j β where q β, i β and, j β 0 and j β = 0 and n β 0 and k t,β 0, f all t n β. In oth cases, solution β implies x, z i β y e which gives either k α = i α k α = i α + j α y 7 and 8, respectively. Equation 4 gives B β y = k α respectively; Case: Assume τ α = 3. Then α is of the shape y a kα q α B β y = k α iα a kα p α q α a j α where p α > q α and i α, j α < k α < i α + j α. Sucase: Assume τ α = 3 and τ β = 6 and β is of the fm q β y where q β. Then u y = w y and the shape of α implies u z = w z and u = w is alanced. Sucase: Assume τ α = 3 and β is not of the previous shape. Then α implies x, zx e and β does not solve e; Case: Assume τ α = 4. Then α is of the shape y a k,α+iα+jα a knα,α+iα+jα a kα q α a k,α+i α+j α a knα,α+iα+jα iα a jα where q α and i α, j α k α < i α + j α and n α 0 and k t,α 0, f all t n α. Solution α implies x iα y, zx k,α z zx knα,α zx kα jα y e. Sucase: Assume τ α = 4 and τ β {2, 3, 5, 8}. Then equation 4 gives A α y A α zb α y = q β + p β + ; a contradiction since gcdp β +, q β + =. Sucase: Assume τ α = 4 and τ β = 6. Then β is of the fm i β a k,β+i β + a k n β,β+i β + a j β

18 8 TERO HARJU AND DIRK NOWOTKA x y a k,β+ a a k qβ n β,β+ a j β where q β, i β and, j β 0 and j β = 0 and n β 0 and k t,β 0, f all t n β. In oth cases, β implies x, z i β y e and k t,α = 0, f all t n α, and k α = j α and equation 4 gives i α q α + j α = B βy B β x ; Case: Assume τ α = 5. Then α is of the shape y a iα+jα+mα q α a j α iα a iα+jα+mα p α where p α, q α, i α, j α and m α 0. Sucase: Assume τ α = 5 and τ β = 2. Then α implies x, zx e and β implies x, z i β y e where i β ; Sucase: Assume τ α = 5 and τ β {3, 4, 8}. Then equations solved y α end with x and y, ut equations solved y β end with x and z y and z; Sucase: Assume τ α = 5 and τ β = 6. Then β is of the fm i β a k,β+i β + a k n β,β+i β + a j β x y a k,β+ a a k qβ n β,β+ a j β where q β, i β and, j β 0 and j β = 0 and n β 0 and k t,β 0, with t β n β. Equation 5 gives q α + p α + = A βy A β xb β y and respectively; a contradiction since gcdp α +, q α + =. Case: Assume τ α = 6. Then α is of the shape i α a k,α+i α+j α a knα,α+iα+jα a j α iα a jα q α + p α + = A βx B β x qα

19 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 9 q α and i α, i α, j α, j α 0 and i α i α = j α j α = 0 and n α 0 and k t,α 0, f all t n α. Sucase: Assume τ α = 6 and τ β {2, 3, 5, 8}. Then equation 4 gives q β + p β + = A αy A α zb α y ; a contradiction since gcdp β +, q β + =. Sucase: Assume τ α = 6 and τ β = 4. Then β is of the shape i β a k,β+i β + a k n β,β+i β + a k qβ β a k,β +i β + a k n β,β+i β + where q β and i β, k β < i β + and n β 0 and k t,β 0, f all t n β. Now, i α = j α = 0 and x iα y, z i β y e and k t,β = 0, f all t n β, and i α = n β +. This implies k β = and k t,α = 0, f all t n α and equation 4 gives q β + n β + = q α i α + j α ; Case: Assume τ α = 8. Then we use arguments similar to case τ α = 5. The previous lemmas and Remark 5 imply the following conclusion. Proposition 4. The intersection of two different entire systems of different type contains only alanced equations Entire Systems of Equal Type. Let τ e the type where α and β are oth taken from, and let ρ: A + A + e an isomphism such that ρa = and ρ = a. Lemma 5. If αx = βx = a then K α K β contains only alanced equations. Proof. We have f type that p α + q α + = w x u x = p β + u z w z q β + and p α = p β and q α = q β otherwise gcdp +, q + > f at least one of the two solutions, ut now α is a permutation of β; F all other types Remark 5 implies that equations solved y α and β egin and end with y and z, and we have only to consider types 2 and 6 further. Case: Assume τ = 2. Then i α = j α = i β = = 0 and equation gives q α + p α + = q β + p β + which implies p α = p β and q α = q β otherwise gcdp +, q + > f at least one of the two solutions. Equations solved y α and β imply yx kα x, zx kα x 2 e and yx k β x, zx k β x 2 e respectively, where x, x 2 {y, z} which gives k α = k β and α is a permutation of β;

20 20 TERO HARJU AND DIRK NOWOTKA Case: Assume τ = 6. Then i α = i α = j α = j α = i β = i β = = j β = 0 and equation gives q α n α + = q β n β + q α n α + = q β n β +. In the latter case we get q α = q β = and n α = n β = 0 and α is a permutation of β; In the fmer case, we have αz = βz = and y, vy e where v {x, z} +. Now, y = v s v, with s 0 and v v, which implies together with B α y = B β y that αy and βy are equally defined, and hence, α is a permutation of β; Lemma 6. If αx = βy = a and τ = 2 then K α K β contains only alanced equations. Proof. Let α e of the shape y a kα q α iα a kα p α a j α where p α, q α, k α and i α, j α 0 and i α + j α k α. Now, β is of the shape 9 i β a k pβ β a z a k β 20 x a k β i β a k pβ β a where p β, q β, k β and i β, 0 and i β + k β. Case: Assume β is of shape 9. Then i α, j α and i β = = 0 since equations solved y α and β egin and end with x and z. By Lemma 4 we have p β > q β. Now, α implies x, zx kα jα y e x, zx kα iα jα z e and β implies x, xy e which gives that k α = j α ; a contradiction since now k α < i α + j α k α. Case: Assume β is of shape 20. If e egins with x and z then i α and i β = 0 and q β > p β y Lemma 4 and α implies x, zx kα jα y e x, zx kα iα jα z e and β impliesx, zy e which gives that k α = j α ; a contradiction since we require k α < i α + j α k α. If e egins with y and z then i β and q α > p α y Lemma 4 and i α = 0 and α implies yx kα, zx kα jα y e yx kα, zx kα jα z e and solution β implies y i β xy k β i β, zy k β x e y i β xy k β i β, zy k β i β z e and in any case k α = j α =, since k β i β. Now, equation 2 gives ut, equation 3 gives p α + q α + = p β + q β + k β p β + i β + k β q β p β + q β + = p β + q β + ;

21 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 2 Lemma 7. If αx = βy = a and τ {3, 4, 5, 8} then K α K β contains only alanced equations. Proof. Equations solved y α egin with x and z whereas equations solved y β egin with x and y y and z; Lemma 8. If x, y e and αx = βy = a and τ = 6 then K α K β contains only alanced equations. Proof. Let α e of the shape i α a k,α+j α a knα,α+jα a j α jα where q α, i α and j α, j α 0 and j α j α = 0 and n α 0 and k t,α 0, f all t n α. Actually, j α and j α = 0, otherwise equation 2 gives A αy B α y = B βz B β x ; So, u = w ends in x and z. Now, β is of the shape 2 22 i β k,β+i β a k n β,β+i β x z a i β a k,β a k qβ n β,β qα where q β, i β, i β and n β 0 and k t,β 0, f all t n β. Case: Assume β is of shape 2. Then equation 3 gives q α n α + = q β k t,β i β = t n β which implies q α = q β = and n α = 0, and equation 2 gives j α i α = n β + which implies j α > i α. Now, α implies x, yx e and β implies x, y i β z e; Case: Assume β is of shape 22. Equation 3 gives B α y = A βx B β x ;

22 22 TERO HARJU AND DIRK NOWOTKA Lemma 9. If x, z e and αx = βy = a and τ = 6 then K α K β contains only alanced equations. Proof. Let α e of the shape k,α+iα+jα a knα,α+iα+jα a j α iα a jα where q α, i α and j α, j α 0 and j α j α = 0 and n α 0 and k t,α 0, f all t n α. Now, β is of the shape qα k,β+ a k n β,β+ a j β x a k,β+ a k qβ n β,β+ a j β where q β and, j β 0 and j β = 0 and n β 0 and k t,β 0, f all t n β. Case: Assume β is of shape 23. Then we have βz < βx y Lemma 4, and hence, q β = and j β = n β = 0 and equation 3 gives q α n α + = ; Case: Assume β is of shape 24. If = 0 then equation 3 gives q α n α + = A βx B β x ; So, and j α = j α = j β = 0 since e ends in y and z. Now, equation 2 gives i α t n α k t,α n α + = q β n β + which implies n α + n β + and equation 3 gives q α n α + = t n β k t,β n β + which implies n α + n β +, and hence, n α = n β and q α = q β = and 25 i α k t,α = k t,β = t n t n

23 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 23 where n = n α = n β. Solution α implies x iα y, z e and β implies xy, z e which gives i α =, and α also implies xy, z e and β implies xy, z e which gives =. From equation 25 follows α: β : n n and By Proposition 2.5 in [7] we get the same generic equation f oth entire systems K α and K β generated y α and β, respectively, namely xy = z n+ and hence, K α and K β are not different; Lemma 20. If y, z e and αx = βy = a and τ = 6 then K α K β contains only alanced equations. Proof. Let α e of the shape k,α+jα a knα,α+jα a j α jα where q α and j α, j α 0 and j α j α = 0 and n α 0 and k t,α 0, f all t n α. Actually, j α and j α = 0 since otherwise equation 2 gives A αy B α y = B βz B β x ; qα So, e ends in x and z. Now, β is of the shape x i β a k,β a k n β,β x i β a k,β+i β a k qβ n β,β+i β where q β, i β, i β, j β and n β 0 and k t,β 0, f all t n β. Case: Assume β is of shape 26. Then equation 3 gives q α n α + = q β i β + = t n β k t,β which implies q α = q β = i β = and n α = k t,β = 0, f all t n β. Now, α implies yx jα, z e and β implies yx n β+, z e which gives j α = n β +, and we have α = ρ β;

24 24 TERO HARJU AND DIRK NOWOTKA Case: Assume β is of shape 27. Then equation 2 gives j α t n α k t,α n α + = q β n β + = which implies n α + n β +, and equation 3 gives q α n α + = i β t n β k t,β n β + which implies n β + n α +, and hence, n α = n β and q α = q β = and 28 j α k t,α = i β k t,β = t n t n where n = n α = n β. Solution α implies yx, z e and β implies y i β x, z e which gives i β =, and α also implies yx jα, z e and β implies yx, z e which implies j α =. From equation 28 follows = α: β : n n and By Proposition 2.5 in [7] we get the same generic equation f oth entire systems K α and K β generated y α and β, respectively, namely yx = z n+ and hence, K α and K β are not different; Lemma 2. If αx = βz = a and τ = 2 then K α K β contains only alanced equations. Proof. Let α e of the shape y a kα q α iα a kα p α a j α where p α, q α, k α and i α, j α 0 and i α + j α k α. Now, β is of the shape i β a k pβ β a y a k β 29 x a k β i β a k pβ β a j 30 β where p β, q β, k β, i β, and i β + k β.

25 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 25 Case: Assume β is of shape 29. Then i α, j α since e must egin and end with x and z. Now, solution α implies x, zx kα jα y e x, zx kα iα jα z e and solution β implies x, z i β y e which gives k α = i α + j α. Equation 4 gives k α = q β + p β + ; Case: Assume β is of shape 30. Then i α = j α = 0 and α implies yx, z e and β implies yz k β i β y, z e yz k β x, z e which gives k β = ; Lemma 22. If αx = βz = a and τ = 3 then K α K β contains only alanced equations. Proof. Let α e of the shape y a kα q α iα a kα p α q α a j α where p α > q α and i α, j α < k α < i α + j α. Now, β must e of the shape i β a k pβ β q β a y a k β where p β > q β and i β, < k β < i β +. Solution α implies x, zx u and β implies x, z i β y u; Lemma 23. If αx = βz = a and τ = 4 then K α K β contains only alanced equations. Proof. Let α e of the shape y a k,α+iα+jα a knα,α+iα+jα a kα q α a k,α+i α+j α a knα,α+iα+jα iα a jα where q α and i α, j α k α < i α + j α and n α 0 and k t,α 0, f all t n α. Now, β must e of the shape i β a k,β+i β + a k n β,β+i β + a k qβ β a k,β +i β + a k n β,β+i β + where q β and i β, k β < i β + and n β 0 and k t,β 0, f all t n β. Solution α implies x iα y, zx k,α z zx knα,α zx kα jα y e and β implies xz k,β x xz k n β,β xz k β y, z i β y e which gives that k t,ζ = 0, f all t n ζ and ζ {α, β}, and k α = j α and k β =. Equation 4 gives q α + i α j α = q β + n β + ;

26 26 TERO HARJU AND DIRK NOWOTKA Lemma 24. If αx = βz = a and τ {5, 8} then K α K β contains only alanced equations. Proof. Equations solved y α end in x and y whereas equations solved y β end in x and z y and z; Lemma 25. If x, y e and αx = βz = a and τ = 6 then K α K β contains only alanced equations. Proof. Let α e of the shape i α a k,α+j α a knα,α+jα a j α jα where q α, i α and j α, j α 0 and j α j α = 0 and n α 0 and k t,α 0, f all t n α. Now, β is of the shape 3 32 k,β+ a k n β,β+ a j β x qα a k,β+ a k qβ n β,β+ a j β where q β and, j β 0 and j β = 0 and n β 0 and k t,β 0, f all t n β. Case: Assume β is of shape 3. Then from αx < αy follows y Lemma 4 that βy < βx, and hence, and j β = 0 and p β = and n β = 0, and so, βy =. But now, equation 5 gives = q α n α + ; Case: Assume β is of shape 32. Then equation 4 gives 33 q α i α + j α j α + = q β n β + t n α k t,α which implies q α = q β = and n β = 0, and equation 5 gives j β = n α + and we have and j β = 0 and e ends with y and z. That implies j α = j α = 0 and since equation 33 gives i α + k t,α = t n α and i α, we have that αy = a i α n α+. Solution α implies x iα z, y e and β implies xz, y e which gives i α =. Solution α implies y, xz nα+ e and β

27 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 27 implies y, xz e which gives that n α + =, and now, α = ρ β; Lemma 26. If x, z e and αx = βz = a and τ = 6 then K α K β contains only alanced equations. Proof. Let α e of the shape k,α+iα+jα a knα,α+iα+jα a j α iα a jα where q α, i α and j α, j α 0 and j α j α = 0 and n α 0 and k t,α 0, f all t n α. Now, β is of the shape qα 34 i β a k,β+i β + a k n β,β+i β + a j β 35 x a i β a k,β + a k qβ n β,β+ a j β where q β, i β, i β and, j β 0 and j β = 0 and n β 0 and k t,β 0, f all t n β. Note, that if β is of shape then 36 x iα y, z l y e where l {i β, i β }. Case: Assume β is of shape 34. From 36 follows that k t,ζ = 0, f all t n ζ and ζ {α, β}, and equations 4 and 5 give q β n β + = q α j α i α j α and q α n α + = q β j β i β respectively. This implies j α, j β > and j α = = 0. Now, e must end in x and y y α and in y and z y β; Case: Assume β is of shape 35. From 36 follows that k t,α = 0, f all t n α, and equation 4 gives q β n β + = q αj α i α j α = and q α = q β = and j α and j α = n β = 0 and also and j β = 0, since e ends in y and z. Equation 5 gives n α + = i β which implies > i β, and we get from 36 that ai β, a i β a βu, βw a i β a i β, a i β a βu, βw which implies i β ; Lemma 27. If y, z e and αx = βz = a and τ = 6 then K α K β contains only alanced equations.

28 28 TERO HARJU AND DIRK NOWOTKA Proof. Let α e of the shape k,α+jα a knα,α+jα a j α jα where q α and j α, j α 0 and j α j α = 0 and n α 0 and k t,α 0, f all t n α. Now, β is of the shape i β a k,β + a k n β,β+ a j β qα x a k,β+i β + a k qβ n β,β+i β + a j β i β a where q β, i β, i β and, j β 0 and j β = 0 and n β 0 and k t,β 0, f all t n β. We have f any shape of β that βz < βy, and hence, αy < αz y Lemma 4. So, q α = and j α = n α = 0 and equation 4 gives j α = B βy B β x ; The previous lemmas imply the following conclusion. Proposition 28. The intersection of two different entire systems of the same type contains only alanced equations. Proposition 29 follows directly from Proposition 4 and 28. Proposition 29. The intersection of two different entire systems contains only alanced equations. 6. Aout Systems of Equations This section contains the main result of this article. It states that an independent system with at least two equations and a nonperiodic solution consists of alanced equations only. Befe this statement is proved, we oserve the following propositions. Proposition 30. If two unalanced equations e and e 2 have a common nonperiodic solution, then they have the same set of periodic solutions. Proof. Let α e a nonperiodic solution of e and e 2. Let e i = u i, w i, and let δ x e i = u i x w i x where x X and i {, 2}. Assume that α is incontractale and αx = a without restriction of generality. Now, B α yδ y e i + B α zδ z e i = 0

29 ON THE INDEPENDENCE OF EQUATIONS IN THREE VARIABLES 29 f i {, 2}, and B αy B α z = δ ze δ y e = δ ze 2 δ y e 2 and we have sδ y e = tδ y e 2 and sδ z e = tδ z e 2 with s, t 0. It is easy to see that also sδ x e = tδ x e 2. Clearly, e and e 2 have the same set of periodic solutions. Proposition 3. If an unalanced equation e and a alanced equation e 2 have a common nonperiodic solution, then every solution of e is a solution of e 2. Proof. It is clear that any periodic solution of e is a solution of e 2 since every periodic solution is a solution f a alanced equation. Let α e a nonperiodic solution of e and e 2. Assume that there exists a nonperiodic solution β of e that is not a solution of e 2. Let α and β e incontractale without restriction of generality. Now, α and β generate two different enire systems since e 2 K α and e 2 K β. But, e K α K β which implies that e is alanced y Proposition 29; The main result of this article follows immediately. Theem 32. If a system of equations has a nonperiodic solution and contains an unalanced equation then it is not independent it is a singleton. Proof. Let S e a system of at least two equations that has a nonperiodic solution α and contains at least one unalanced equation e. Assume that S is independent, then it contains no alanced equation y Proposition 3. Let e 2 e an unalanced equation in S different from e. Since S is independent, there exists a solution β that solves e ut does not solve e 2. From Proposition 30 follows that β is nonperiodic. We can assume that oth α and β are incontractale without restriction of generality. Furtherme, α and β generate two different entire systems since e 2 K α and e 2 K β. But, e K α K β which implies that e is alanced y Proposition 29; Collary 33. An independent system with at least two equations and a nonperiodic solution consists f alanced equations only. 7. Conclusions We have shown that the intersection of the kernel of two different nonperiodic solutions in three variales contains only alanced equations, Proposition 29. From that result and the fact that the independence of systems in three variales depends on their nonperiodic solutions only, Proposition 30 and 3, follows that independent systems of equations in three variales that have a nonperiodic solution and contain me than one equation consist of alanced equations only, Theem 32 and Collary 33. This result is a further step towards an answer of the question whether not an independent system of three equations in three variales with a nonperiodic solution exists. We have estalished here that length arguments do not help to answer that question.

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