aa r X i v : . [ m a t h . G M ] O c t Identities of the Function f ( x, y ) = x + y Abstract
Harvey Friedman asked in 1986 whether the function f ( x, y ) = x + y on thereal plane R satisfies any identities; examples of identities are commutativity andassociativity. To solve this problem of Friedman, we must either find a nontrivialidentity involving expressions formed by recursively applying f to a set of variables { x , x , . . . , x n } that holds in the real numbers or to prove that no such identitieshold. In this paper, we will solve certain special cases of Friedman’s problem andexplore the connection between this problem and certain Diophantine equations. Acknowledgements
I would like to thank my thesis advisor, George Bergman, for introducing Lemma2 to me, shortening the proofs of several of the results below, pointing out areasof the paper that needed clarification, and giving advice on how to better organizethis thesis.
Notice that, given any identity in any number of variables, one can get an identityin one variable x by replacing all the variables of the given identity by x . The singlevariable case of Friedman’s problem, whether or not there exists a nontrivial identitythat holds in the real numbers involving expressions formed by recursively applying f to the variable set { x } , may be easier to treat than the general problem of multiplevariables. A priori , proving that no nontrivial identity of one variable holds does not com-pletely solve the general problem, because two expressions, if equal as polynomi-als, that have the same “structure” regarding the composition of f ’s (ignoring thevariables involved) lead to the trivial identity when all the variables are replacedby x . For instance, f ( f ( x, y ) , f ( y, x )) = f ( f ( y, x ) , f ( x, y )) as polynomials implies f ( f ( x, x ) , f ( x, x )) = f ( f ( x, x ) , f ( x, x )) as polynomials. However, proving that no non-trivial identity of one variable holds would tell us that two expressions can be equalas polynomials only if they have the same structure. For instance, since f ( x, f ( x, x ))and f ( f ( x, f ( x, x )) , x ) are not equal as polynomials, we know that f ( x, f ( y, z )) and f ( f ( x, f ( y, z )) , y ) cannot be equal as polynomials. We will use this observation toprove in Lemma 2 that a nontrivial multiple-variable identity holds only if a nontrivial1-variable identity holds.We will follow the convention that 0 / ∈ N .1 otation . Suppose G ( x , x , . . . , x n ) is an expression formed by recursively apply-ing f to the variable set { x , x , . . . , x n } . We shall call an occurrence of a variablein G ( x , x , . . . , x n ) a variable position . Supposing that G ( x , x , . . . , x n ) contains l variable positions where l ∈ N , we will proceed from left to right and label thesesuccessive variable positions as v , v , . . . , v l , and for all i = 1 , , . . . , l we will de-note by ¯ v i the variable in { x , x , . . . , x n } occurring in the variable position v i . Wedefine the depth of a variable position v occurring in an f -expression f ( A, B ) tobe one more than its depth in A or B (whichever v occurs in), where we start bydefining the depth of the bare expression x i where i ∈ { , , . . . , n } to be 0; we willdenote the depth of v by depth( v ). For example, the variable positions v , v , v of f ( x, f ( y, x )) hold the variables ¯ v = x , ¯ v = y , and ¯ v = x , while we have depth( v ) = 1,depth( v ) = 2 and depth( v ) = 2. We can associate to G ( x , x , . . . , x n ) the l -tuple(( ¯ v , depth( v )) , ( ¯ v , depth( v )) , . . . , ( ¯ v l , depth( v l ))). It is clear that G ( x , x , . . . , x n )completely determines (( ¯ v , depth( v )) , ( ¯ v , depth( v )) , . . . , ( ¯ v l , depth( v l ))).The following result was pointed out to the author by George Bergman, and itsproof follows the ideas outlined by Bergman. This lemma establishes that, to answerFriedman’s problem in the negative, it suffices to prove that no nontrivial identity ofone variable holds. Lemma 2.
Suppose that f satisfies no nontrivial 1-variable identities in R . Let n bea positive integer and let { x , x , . . . , x n } be a set of variables. Then f satisfies nonontrivial identities involving the variables x , x , . . . , x n in R .Proof. Suppose that two distinct expressions G ( x , x , . . . , x n ), H ( x , x , . . . , x n )formed by recursively applying f to { x , x , . . . , x n } are equal as polynomials. It followsthat G ( x, x, . . . , x ) and H ( x, x, . . . , x ) are equal as polynomials. Then, by the assump-tion of no nontrivial 1-variable identities, G ( x, x, . . . , x ) and H ( x, x, . . . , x ) must be thesame expression, so G ( x , x , . . . , x n ) and H ( x , x , . . . , x n ) must have the same numberof variable positions. We will label the variable positions of G ( x , x , . . . , x n ) by v , v ,. . . , v l and the variable positions of H ( x , x , . . . , x n ) by v ′ , v ′ , . . . , v ′ l , where l is somepositive integer. Now, G ( x , x , . . . , x n ) determines the l -tuple(( ¯ v , depth( v )) , ( ¯ v , depth( v )) , . . . , ( ¯ v l , depth( v l )))and H ( x , x , . . . , x n ) determines the l -tuple(( ¯ v ′ , depth( v ′ )) , ( ¯ v ′ , depth( v ′ )) , . . . , ( ¯ v ′ l , depth( v ′ l ))) . We know that for each i = 1 , , . . . , l we have depth( v i ) = depth( v ′ i ), i.e. correspondingvariable positions have the same depth. Let j be the smallest positive integer such that¯ v j = x k = x m = ¯ v ′ j where k = m . Now replace x k in G ( x , x , . . . , x n ), H ( x , x , . . . , x n )by f ( x, x ) and replace x i in G ( x , x , . . . , x n ), H ( x , x , . . . , x n ) by x for each i = k ,and we obtain a 1-variable identity. Then there exists at least one p ≥ j in N such thatthe p th variable position of the 1-variable expression resulting from G ( x , x , . . . , x n )has a depth one greater than that of the p th variable position of the 1-variable expres-sion resulting from H ( x , x , . . . , x n ). Therefore, the two 1-variable expressions in theidentity are distinct, which is a contradiction.2he proof of Lemma 2 leaves open a more difficult question, as the statement ofLemma 2 is weaker than what we state in the following Conjecture 3.
Suppose that G ( x ) is an expression formed by recursively applying f tothe variable set { x } and that G ( x ) has the variable positions v , v , . . . , v l for somepositive integer l . Let n be a positive integer and { x , x , . . . , x n } be a set of variables.Let G ( x , x , . . . , x n ) be an n-variable expression obtained by letting ¯ v i ∈ { x , x , . . . , x n } for all i = 1 , , . . . , l . If G ( x ) cannot occur as either side of a nontrivial 1-variableidentity, then G ( x , x , . . . , x n ) cannot occur as either side of a nontrivial n -variableidentity. Below, we prove some results on the single variable case of Friedman’s problem.They show that certain classes of expressions cannot occur as either side of a nontrivialidentity.
Definition 4. If p ( x ) = P nk = m a k x k is a polynomial where m ≤ n are nonnegativeintegers and where a m , a n are nonzero, then m will be called the order of p ( x ) , and n will be called the degree of p ( x ) . In what follows, by an f -expression we will, unless otherwise specified, alwaysmean a symbolic expression in f and x that is formed by recursively applying f to thevariable set { x } ; we also consider x itself an f -expression. We will denote the set of all f -expressions by term( f ; x ). Let e : term( f ; x ) −→ Z [ x ] be the evaluation map thatassigns to each f -expression its corresponding polynomial in Z [ x ]. We say that e ( A )is the polynomial induced by the f -expression A . For example, e ( f ( x, f ( x, x ))) = x + ( x + x ) . If A and B are two f -expressions and e ( A ) = e ( B ), then we saythat A and B are e -equivalent . We shall call an f -expression e -isolated if it isnot e -equivalent to any other f -expression, i.e. it cannot occur as either side of anontrivial identity. For example, f ( x, x ) is e -isolated because, as it is not hard to see, e ( f ( A , A )) = e ( f ( x, x )), where A , A are f -expressions, implies A = x and A = x . Notation . Let A ∈ term( f ; x ). For brevity, we will denote the degree of e ( A ) bydege( A ) and the order of e ( A ) by orde( A ). For the degree and order of any polynomial p ( x ) that is not written as the induced polynomial of some B ∈ term( f ; x ), we retain thestandard notation deg( p ( x )) and ord( p ( x )) respectively. For example, we would denotethe degree of p ( x ) = x + x as deg( p ( x )) if we did not explicitly state or did not knowbeforehand that p ( x ) = e ( f ( x, x )).In the next three propositions, let A, B ∈ term( f ; x ). Proposition 6. If e ( f ( C , C )) = e ( f ( x, B )) where C , C are f -expressions, then wemust have C = x and e ( C ) = e ( B ) .Proof. We have e ( f ( x, B )) = x + e ( B ) = e ( C ) + e ( C ) . Notice that 3 orde( C ) ≥ x must arise in the expansion of e ( C ) . This forces C = x , because otherwiseevery term that arises in the expansion of e ( C ) will have powers at least as high as 4.Then cancellation gives us e ( C ) = e ( B ) , which forces e ( C ) = e ( B ).3 roposition 7. If e ( f ( C , C )) = e ( f ( A, x )) where C , C are f -expressions, then wemust have e ( C ) = e ( A ) and C = x .Proof. We have e ( f ( A, x )) = e ( A ) + x = e ( C ) + e ( C ) . If C = x , then e ( C ) = x .If C = x , then orde( C ) ≥
2, so e ( C ) has order at least 4. Thus, x must arise in theexpansion of e ( C ) . Therefore, we have C = x and it follows from cancellation that e ( C ) = e ( A ) , so e ( C ) = e ( A ). Remark . Actually, the arguments in the proofs of the previous two propositions alsoapply if f ( x, B ) in Proposition 6 and f ( A, x ) in Proposition 7 are instead multiple-variable f -expressions. For example, the same arguments can be applied, repeatedly, toshow that f -expressions such as f ( f ( x, f ( y, z )) , y ) are e -isolated. In effect, this settlesa special case of Conjecture 3. Proposition 9. If e ( f ( C , C )) = e ( f ( f ( x, x ) , B )) where C , C are f -expressions,then we must have C = f ( x, x ) and e ( C ) = e ( B ) .Proof. We have e ( C ) + e ( C ) = e ( f ( x, x )) + e ( B ) = ( x + x ) + e ( B ) = x +2 x + x + e ( B ) . If B = x , then the conclusion follows by Proposition 7. Suppose B isnot x . Notice that the terms x and 2 x must arise in the expansion of e ( C ) because3 orde( C ) ≥ C = x . Since x = x · x , we must have e ( C ) = e ( f ( x, C )) = x + e ( C ) where C is another f -expression. Considering ( x + e ( C ) ) and the fact that x = x · x show that the expansion of e ( C ) contains the term x , so e ( C ) = x andthus C = x . Since e ( C ) = x + x , it follows again by cancellation that e ( C ) = e ( B ) and so e ( C ) = e ( B ). Since f ( x, x ) is e -isolated, we have C = f ( x, x ).As can be seen, the above three propositions were established with an argument thatworks “outside-in” in the sense that it depends only on the x and f ( x, x ) that are beingappended to A , B by f , while A and B can be completely arbitrary. This argument isdifficult to apply for f -expressions such as f ( C, B ), where B is arbitrary and C is an f -expression such that dege( C ) > dege( f ( x, x )). Below we will introduce an argumentthat works, in some sense, “inside-out.”Let f ( A, B ) be an f -expression. We will be examining the leading terms of thepolynomial e ( f ( A, B )) by looking at the subexpressions from which they arise. Forinstance, e ( f ( x, f ( x, x ))) = x + e ( f ( x, x )) = x + ( x + x ) and we see that the termwith the degree of the polynomial arises from the f ( x, x ) by the product ( x ) = x .The next lemma will show that the fact that this highest degree term arises from asubexpression f ( x, x ) is very generally true. Lemma 10.
Every summand contributing to the highest degree term of e ( f ( A, B )) mustarise from an occurrence of f ( x, x ) contained in f ( A, B ) , on expanding e ( f ( A, B )) inpowers of x .Proof. Consider the polynomial expansion of e ( f ( A, B )). Suppose dege( f ( A, B )) =2 m n for some m, n ∈ N ∪ { } . Then the highest degree term of e ( f ( A, B )) is px m n forsome p ∈ N . Let x d denote an occurrence of x in f ( A, B ) such that at least one of the4 copies of x m n (we will denote this copy by [ x m n ] α ) in the expansion of e ( f ( A, B ))contains at least one factor of this occurrence of x , i.e. [ x m n ] α = x d · x m n − or[ x m n ] α = x d · x m n − . Suppose that x d is contained in a subexpression f ( x d , C ) or f ( D, x d ) where C = x and D = x . Considering the product e ( C ) · x m n − that arisesfrom f ( x d , C ) and the product e ( D ) · x m n − that arises from f ( D, x d ), we see that inthe expansion of e ( f ( A, B )) any occurrence of x contained in C or in D will lead to aterm with a power higher than 2 m n . This is a contradiction, so x d must be containedin an f ( x, x d ) in f ( A, B ).We will call an occurrence of f ( x, x ) in the f -expression f ( A, B ) a core of f ( A, B )if this occurrence of f ( x, x ) gives rise to a summand contributing to the highest degreeterm of e ( f ( A, B )). Whenever an occurrence f ( x, x ) ′ of f ( x, x ) in f ( A, B ) is a core of f ( A, B ), it is clear that e ( f ( x, x ) ′ ) dege( f ( A,B )) = ( x + x ) dege( f ( A,B )) must be a termof e ( f ( A, B )).We define inductively what it means to develop an f -expression about a core:1. Start with f ( x, x ) and label it a core of the f -expression to be developed. Then f ( x, x ) is the f -expression at the first stage of the development.2. Let A be the f -expression at the n th stage of the development where n ≥
1. Thenthe f -expression at the ( n + 1)st stage of the development is either f ( A, C ) where C is an f -expression such that 3 dege( C ) ≤ A ) or f ( C, A ) where C is an f -expression such that 2 dege( C ) ≤ A ).Any f -expression can be developed inductively in the above manner, though the de-velopment may not be unique. For example, we can develop f ( x, f ( x, x )) only by the se-quence of steps f ( x, x ), f ( x, f ( x, x )) while we can develop f ( f ( x, f ( x, x )) , f ( f ( x, x ) , x ))by the sequence f ( x, x ), f ( x, f ( x, x )), f ( f ( x, f ( x, x )) , f ( f ( x, x ) , x )) or by the sequence f ( x, x ), f ( f ( x, x ) , x ), f ( f ( x, f ( x, x )) , f ( f ( x, x ) , x )). However, every development of an f -expression whose induced polynomial has degree 2 m n must consist of m + n stages.Suppose that C and C are distinct cores of the f -expression A . Then we can find asubexpression f ( D , D ) of A such that either D contains C and D contains C or viceversa. Since C , C are both cores, we must have D = D and 2 dege( D ) = 3 dege( D ).Since f ( D , D ) is the f -expression at the n th stage of the development of A about C and the f -expression at the n th stage of the development of A about C for some n ∈ N ,we see that the development of A about C and the development of A about C differ atthe ( n − f -expression has a unique core whenever it has a unique development. Therefore,an f -expression has a unique development if and only if this f -expression has a uniquecore. Note that a “development” is defined as a property of an f -expression, not ofa polynomial. So far as we know, for a given f -expression, the properties of having aunique core and of being e -isolated are independent of one another.It is easy to see that an f -expression corresponding to a non-monic polynomialdoes not have a unique core. However, the number of cores of an f -expressiondo not necessarily equal the leading coefficient of the induced polynomial; consider5 ( x, f ( f ( x, f ( x, x )) , f ( f ( x, x ) , x ))), which has two cores while the polynomial it induceshas a leading coefficient of 8. Moreover, it is easy to prove by induction the following Lemma 11.
Suppose A ∈ term( f ; x ) and e ( A ) is non-monic. Then we have dege( A ) =2 p q where p ≥ and q ≥ .Proof. Let A = f ( B, C ). If 2 dege( B ) = 3 dege( C ), then neither B nor C is x , so thevalue of dege( f ( B, C )) = 2 dege( B ) = 3 dege( C ) will be divisible by 2 and 3 , the latterbecause dege( C ) is divisible by 3. If 2 dege( B ) = 3 dege( C ), then whichever of B or C contributes the higher degree term must be non-monic, and in that case we may assumeinductively that either dege( B ) or dege( C ) satisfies the conclusion of the lemma.We shall call an f -expression f ( A, B ) disjoint if 2 dege( A ) < B ) or if3 dege( B ) < A ). f ( A, B ) is called hereditarily disjoint if it is disjoint atevery stage of its development about some core. We also consider x to be (vacuously)hereditarily disjoint. Proposition 12. An f -expression A is hereditarily disjoint if and only if it is either1. x f ( x, U ) for U a hereditarily disjoint f -expression3. f ( U, x ) for U a hereditarily disjoint f -expression with orde( U ) ≥ f ( f ( x, x ) , U ) for U a hereditarily disjoint f -expression with orde( U ) ≥ In these four cases, orde( A ) = 1 , , , respectively, and it is easy to see that A has aunique core in all cases except A = x .Proof. If A belongs to one of the above four cases, then A is hereditarily disjoint bydefinition. Now assume that A is hereditarily disjoint, we will show that A belongs toone of the above four cases. Suppose that A = x and that for each core there are n stages in the development of A about that core. Suppose that for all i ≤ n − f -expression at the i th stage of the development of A about each of its cores belongsto one of the above four cases. Then either A = f ( B, C ) where 3 orde( C ) > B )or A = f ( C, B ) where 2 orde( C ) > B ). By the inductive hypothesis, we haveorde( C ) ≤
4, which forces dege( B ) < A = f ( B, C ) and dege( B ) < A = f ( C, B ). Thus, A = f ( B, C ) implies B = f ( x, x ) or B = x , and A = f ( C, B ) implies B = x . This completes the induction. Notation . Let
A, B ∈ term( f ; x ). Whenever we denote A by f ( . . . B . . . ), we meanthat B is a subexpression of A and B contains a core of A .Suppose A := f ( . . . f ( C, B ) . . . ) is an f -expression of degree 2 p q where B containsa core of A . Suppose 3 dege( B ) = 2 m n and 2 dege( C ) = 2 i j . Define the degree-gapbetween C and B to be the positive integerdgap( C, B ) := 3 dege( B ) − C ) = 2 m n − i j . (1)6n the dgap( − , − ) notation we use, we will ignore the order of B and C . In other words,we could also have written (1) as “dgap( B, C ) := 3 dege( B ) − C ) = 2 m n − i j ”(as we have already specified that B contains a core of A ). Now consider the expansionof ( e ( C ) + e ( B ) ) p − m q − n and notice that the highest degree monomial which cancontain a factor coming from C in the expansion of e ( A ) ismaxt A ( C ) = x i j · ( x m n ) p − m q − n − = x i j +(2 m n )(2 p − m q − n − ; (2)here we are ignoring the coefficient of x i j +(2 m n )(2 p − m q − n − , as it is irrelevant at thispoint. The definition in (2) is relative to A given, but we will abbreviate maxt A ( C ) tomaxt( C ) where there is no danger of confusion. Of course, B gives rise to the highestdegree monomial x p q = x dege( A ) of A . Notice thatdege( A ) − deg(maxt( C )) = deg(( e ( B ) ) p − m q − n ) − deg(maxt( C ))= 2 p q − (2 i j + (2 m n )(2 p − m q − n − m n − i j = dgap( C, B ) , so the degree-gap between C and B is preserved in the expansion of e ( A ). This(and its analogue in the next paragraph) will be an important fact in Lemma 17 andProposition 22, where we will prove that an f -expression is e -isolated by considering allpossible developments that lead to an f -expression e -equivalent to the given one.Similarly, in the opposite case, where B := f ( . . . f ( A, D ) . . . ) is an f -expression ofdegree 2 p q , A contains a core of B , 2 dege( A ) = 2 m n , and 3 dege( D ) = 2 i j , we candefine the degree-gap between A and D to be the positive integerdgap( A, D ) := 2 dege( A ) − D ) (3)and notice that the highest degree monomial which can contain a factor comingfrom D in the expansion of e ( B ) ismaxt B ( D ) = x i j +(2 m n )(2 p − m q − n − . (4)Again, we will ignore the order of A and D in the dgap( − , − ) notation we use, and wewill abbreviate maxt B ( D ) to maxt( D ) where there is no danger of confusion. As before,we can observe that dege( B ) − deg(maxt( D )) = dgap( A, D ) . For an f -expression A := f ( . . . B . . . ) where deg( e ( B )) = 2 m n − , we say that B is e -isolated with respect to A if, for every development of every f -expression e -equivalent to A , we obtain the f -expression B (not merely some f -expression e -equivalent to B ) at the ( m + ( n − B is e -isolated with respect to A , then B must be e -isolated. The converse is not true,because even though f ( x, f ( x, x )) is e -isolated, it is not e -isolated with respect to f ( f ( x, f ( x, x )) , f ( f ( x, x ) , x )). Thus, B being e -isolated with respect to A is a strongerstatement than B being e -isolated. Note also that Lemma 10 is equivalent to the state-ment that f ( x, x ) is e -isolated with respect to every f -expression other than x .7 efinition 14. Let A ∈ term( f ; x ) and let D ( A ) , D ( A ) denote two developments of A , not necessarily distinct. We shall say that D ( A ) , D ( A ) agree at the n th stageif there exists ˆ A ∈ term( f ; x ) such that ˆ A is the f -expression at the n th stage of both D ( A ) and D ( A ) .Notation . Given A ∈ term( f ; x ), we will write A [ n ] for the f -expression at the n thstage of the development of A , provided that all developments of A agree at the n thstage. Note that this notation refers only to developments of A , and not to developmentsof f -expressions e -equivalent to A , in contrast to the definition of “ e -isolated with respectto A ” in the paragraph preceding Definition 14. Definition 16.
Let p ( x ) = a n x n + a n − x n − + . . . + a x + a and q ( x ) = b n x n + b n − x n − + . . . + b x + b be two polynomials with nonnegative coefficients. Suppose p ( x ) = q ( x ) , and let m be the greatest integer such that a m = b m . Then we say that p ( x ) is lexicographically greater than q ( x ) , denoted by p ( x ) > L q ( x ) , if a m > b m . Lemma 17.
Let A = f ( . . . f ( x ′ , B ) . . . ) be an f -expression where x ′ := x for the purposeof distinguishing it from the other occurrences of the variable x in A , dege( A ) = 2 p q ,and B contains a core of A . Suppose that for every f -expression C in the ellipses ( . . . ) of A we have deg(maxt( C )) ≤ deg(maxt( x ′ )) . Suppose there exists either an f -expression ¯ A = f ( . . . f ( U, B ) . . . ) such that dege( ¯ A ) = 2 p q , U = x , and B contains a core of ¯ A oran f -expression ˆ A = f ( . . . f ( B, V ) . . . ) such that dege( ˆ A ) = 2 p q and B contains a coreof ˆ A . Then e ( A ) < L e ( ¯ A ) if ¯ A exists and e ( A ) < L e ( ˆ A ) if ˆ A exists.Proof. By our assumption, the subexpressions in the ellipses of A give rise to termswith powers no higher than that of maxt( x ′ ). Suppose e ( B ) is of degree 2 m n .Notice that ( e ( B ) ) p − m q − n = ( e ( B ) ) p − m − q − n +1 is common to both e ( A ) and e ( ¯ A ) (if ¯ A exists), and is common to both e ( A ) and e ( ˆ A ) (if ˆ A exists). Wehave deg( e ( A ) − ( e ( B ) ) p − m q − n ) = deg(maxt( x ′ )), deg( e ( ¯ A ) − ( e ( B ) ) p − m q − n ) ≥ deg(maxt( U )), and deg( e ( ˆ A ) − ( e ( B ) ) p − m − q − n +1 ) ≥ deg(maxt( V )). Since U = x ,we have deg(maxt( U )) > deg(maxt( x ′ )), so deg( e ( A ) − ( e ( B ) ) p − m q − n ) < deg( e ( ¯ A ) − ( e ( B ) ) p − m q − n ). It follows that e ( A ) < L e ( ¯ A ) (in the case that ¯ A exists) as desired.Since dgap( x ′ , B ) = 2 m n − > m +1 n − − ≥ m +1 n − − V ) = dgap( B, V ),we have deg(maxt( x ′ )) = 2 p q − dgap( x ′ , B ) < p q − dgap( B, V ) = deg(maxt( V )),so deg( e ( A ) − ( e ( B ) ) p − m q − n ) < deg( e ( ˆ A ) − ( e ( B ) ) p − m − q − n +1 ). It follows that e ( A ) < L e ( ˆ A ) (in the case that ˆ A exists) as desired. Lemma 18.
Let A = f ( . . . f ( B, x ′ ) . . . ) be an f -expression such that x ′ := x , dege( A ) =2 p q , and B contains a core of A . Suppose ¯ A = f ( . . . f ( B, U ) . . . ) is an f -expressionsuch that U = x , dege( ¯ A ) = 2 p q , and B contains a core of ¯ A . Suppose that for every f -expression C in the ellipses ( . . . ) of A we have deg(maxt( C )) ≤ deg(maxt( x ′ )) . Then e ( A ) < L e ( ¯ A ) .Proof. By our assumption, the subexpressions in the ellipses of A give rise to termswith powers no higher than that of maxt( x ′ ). Suppose e ( B ) is of degree 2 m n . No-tice that ( e ( B ) ) p − m q − n is common to both e ( A ) and e ( ¯ A ). We have deg( e ( A ) − e ( B ) ) p − m q − n ) = deg(maxt( x ′ )) and deg( e ( ¯ A ) − ( e ( B ) ) p − m q − n ) ≥ deg(maxt( U )).Since U = x , we have deg(maxt( U )) > deg(maxt( x ′ )), so deg( e ( ¯ A ) − ( e ( B ) ) p − m q − n ) > deg( e ( A ) − ( e ( B ) ) p − m q − n ). It follows that e ( A ) < L e ( ¯ A ) as desired. Lemma 19.
Let U ∈ term( f ; x ) where U = x . Let B (1) U = f ( x , U ) where x := x . For every positive integer n , let B ( n +1) U = f ( x n +1 , B ( n ) U ) , where x n +1 := x . Let A = f ( . . . B ( n ) U . . . ) where n ∈ N . Then deg(maxt( x )) > deg(maxt( x )) > . . . > deg(maxt( x n )) .Proof. Since B ( n ) U contains a core of A , U must contain a core of A . No-tice that dgap( x , U ) < dgap( x , B (1) U ) < . . . < dgap( x n , B ( n − U ). Sincedeg(maxt( x )) = dege( A ) − dgap( x , U ), deg(maxt( x )) = deg( A ) − dgap( x , B (1) U ),. . . , and deg(maxt( x n )) = dege( A ) − dgap( x n , B ( n − U ), the conclusion immediately fol-lows.Analogously, we have the following Lemma 20.
Let V ∈ term( f ; x ) where V = x . Let C (1) V = f ( V, x ) where x := x . For every positive integer n , let C ( n +1) V = f ( C ( n ) V , x n +1 ) , where x n +1 := x . Let A = f ( . . . C ( n ) V . . . ) where n ∈ N . Then deg(maxt( x )) > deg(maxt( x )) > . . . > deg(maxt( x n )) . Lemma 21.
Let A = f ( . . . f ( f ( x , B ) , x ) . . . ) be an f -expression such that x := x , x := x and B = x . Then deg(maxt( x )) > deg(maxt( x )) .Proof. Notice that B must contain a core of A . We have dgap( x , B ) = 3 dege( B ) − < B ) − f ( x , B ) , x ). Since deg(maxt( x )) = dege( A ) − dgap( x , B ) anddeg(maxt( x )) = dege( A ) − dgap( f ( x , B ) , x ), the conclusion immediately follows. Proposition 22.
Let A = f ( . . . f ( x ′ , B ) . . . ) be an f -expression such that x ′ := x .Suppose that for every f -expression C in the ellipses ( . . . ) of A we have deg(maxt( C )) ≤ deg(maxt( x ′ )) . Suppose that B is e -isolated with respect to A . Then f ( x ′ , B ) is e -isolated with respect to A .Proof. Suppose 3 dege( B ) = 2 m n . Let A ′ be an f -expression e -equivalent to A . Then A ′ [ m +( n − = B by our assumption, so the ( m + n )th stage of every developmentof A ′ is either f ( U, B ) or f ( B, V ) for some
U, V ∈ term( f ; x ). Suppose that either A ′ = f ( . . . f ( U, B ) . . . ) where U = x or A ′ = f ( . . . f ( B, V ) . . . ). By Lemma 17 we have e ( A ) < L e ( A ′ ), which is a contradiction. It follows that the ( m + n )th stage of everydevelopment of A ′ must be of the form f ( U, B ) where U = x . Hence A ′ [ m + n ] = f ( x, B )as desired. Notation . Let B (1) = f ( x, x ). For every positive integer n , let B ( n +1) = f ( x, B ( n ) ).Also, we let B (0) = x . 9otation 23 is to be used for the remainder of this paper, and is not to be confusedwith the ad hoc notations set up in Lemma 19 and Lemma 20. Corollary 24.
Let A = f ( . . . f ( x ′ , B ( m ) ) . . . ) be an f -expression where x ′ := x . Sup-pose that for every f -expression C in the ellipses ( . . . ) of A we have deg(maxt( C )) ≤ deg(maxt( x ′ )) . Then f ( x ′ , B ( m ) ) is e -isolated with respect to A .Proof. We know that B (1) = f ( x, x ) is e -isolated with respect to A . Suppose weknow that B ( k ) is e -isolated with respect to A for some 1 ≤ k ≤ m . Writing A as f ( . . . f ( x ′′ , B ( k ) ) . . . ) where x ′′ := x , we see by Lemma 19 that for every f -expression D in the ellipses of f ( . . . f ( x ′′ , B ( k ) ) . . . ) we have deg(maxt( D )) ≤ deg(maxt( x ′′ )). Itthen follows by Proposition 22 that f ( x ′′ , B ( k ) ) is e -isolated with respect to A . Thiscompletes the induction. Lemma 25.
Let A = f ( . . . f ( x ′ , B ( m ) ) . . . ) be an f -expression where x ′ := x . Let ¯ A = f ( . . . C . . . ) be an f -expression such that dege( ¯ A ) = dege( A ) , C = f ( x ′ , B ( m ) ) , and dege( C ) = dege( f ( x ′ , B ( m ) )) . Suppose that for every f -expression D in the ellipses of A we have deg(maxt( D )) ≤ deg(maxt( x ′ )) . Then e ( A ) < L e ( ¯ A ) .Proof. C has a unique core by Lemma 11. Since C = f ( x ′ , B ( m ) ) and dege( C ) =dege( f ( x ′ , B ( m ) )), we can find some k ≤ m in N such that C [ k +1] = f ( U, B ( k ) ) where U = x . Then we can write ¯ A as f ( . . . f ( U, B ( k ) ) . . . ). Since f ( x ′ , B ( m ) ) [ k +1] = f ( x ′′ , B ( k ) )where x ′′ := x , we can write A as f ( . . . f ( x ′′ , B ( k ) ) . . . ). We see by Lemma 19 that forevery f -expression E in the ellipses of f ( . . . f ( x ′′ , B ( k ) ) . . . ) we have deg(maxt( E )) ≤ deg(maxt( x ′′ )). It follows by Lemma 17 that e ( A ) < L e ( ¯ A ).Lexicographic ordering on polynomials with nonnegative integer coefficients is awell-ordering. In particular, for all m and n the set of all polynomials of degree 2 m n induced by f -expressions contains exactly one lexicographically minimal polynomial,and we will see that this polynomial corresponds to an e -isolated f -expression.As an illustration, we claim that the f -expression f ( f ( f ( x, f ( x, x )) , x ) , x ) leads tothe lexicographically minimal polynomial with degree 36 = (2 )(3 ). The followinglemma gives the general rule. Lemma 26.
For any A ∈ term( f ; x ) , let u ( A ) = f ( x, A ) and v ( A ) = f ( A, x ) . The f -expression that induces the lexicographically minimal polynomial of degree m n with n ≥ is v ( . . . v ( u ( . . . u ( x ) . . . )) . . . ) with m v ’s followed by n u ’s in left-to-right order.Moreover, this f -expression is e -isolated.Proof. Let A := v ( . . . v ( u ( . . . u ( x ) . . . )) . . . ) with m v ’s followed by n u ’s in left-to-rightorder, and let A ′ ∈ term( f ; x ) be such that A ′ = A and dege( A ′ ) = dege( A ). It isclear that A has exactly one core. Let k be the largest positive integer such that thedevelopment of A agrees with every development of A ′ at the k th stage. Let C denotethe f -expression at the k th stage of the development of A . Suppose k < n . Then wecan write A as f ( . . . f ( x , C ) . . . ) and we can write A ′ as either f ( . . . f ( U, C ) . . . ) or f ( . . . f ( C, V ) . . . ) where x := x and U, V ∈ term( f ; x ) such that U = x . By Lemma109, Lemma 20, and Lemma 21, we have deg(maxt( x )) > deg(maxt( D )) for every f -expression D in the ellipses of f ( . . . f ( x , C ) . . . ). It follows by Lemma 17 that e ( A ) < L e ( A ′ ). Suppose k ≥ n . Then we can write A as f ( . . . f ( C, x ) . . . ) and we can write A ′ as f ( . . . f ( C, W ) . . . ), where x := x and W ∈ term( f ; x ) such that W = x . ByLemma 20, we have deg(maxt( x )) > deg(maxt( E )) for every f -expression E in theellipses of f ( . . . f ( C, x ) . . . ). It follows by Lemma 18 that e ( A ) < L e ( A ′ ). Thus, inall cases we have e ( A ) < L e ( A ′ ). It immediately follows from this analysis that A is e -isolated, though we could well have proven this particular fact by repeatedly applyingProposition 6 and Proposition 7.As we will see, this concept of lexicographic minimality can be applied to prove thatmany classes of f -expressions are e -isolated. This concept also illustrates one advantageof working with the single-variable case of Friedman’s problem, because it is less clearhow one would lexicographically order multiple-variable polynomials. Proposition 27.
Let f ( f ( F, x ′ ) , B ) be an f -expression such that e ( f ( f ( F, x ′ ) , B )) hasdegree m n , where x ′ := x . Suppose that1. B ) ≤ deg(maxt( x ′ )) , where we note that deg(maxt( x ′ )) = 3 + 2 dege( F ) e ( F ) is lexicographically minimal among polynomials of degree m − n induced by f -expressions as characterized in Lemma 263. a , b are f -expressions such that e ( f ( a, b )) = e ( f ( f ( F, x ′ ) , B )) .Then we must have a = f ( F, x ′ ) and e ( b ) = e ( B ) .Proof. First notice that m ≥ e ( f ( F, x ′ )) is lexicographi-cally minimal among polynomials of degree 2 m − n induced by f -expressions. Writ-ing f ( f ( F, x ′ ) , B ) as f ( . . . f ( x ′′ , B ( n − ) . . . ) where x ′′ := x , we see by Lemma 20 andLemma 21 that deg(maxt( C )) ≤ deg(maxt( x ′′ )) for every f -expression C in the el-lipses of f ( . . . f ( x ′′ , B ( n − ) . . . ). It follows by Corollary 24 that f ( a, b ) [ n ] = B ( n ) = f ( f ( F, x ′ ) , B ) [ n ] . Assume we know that f ( a, b ) [ k ] = f ( f ( F, x ′ ) , B ) [ k ] for some posi-tive integer n ≤ k < m − n . Suppose f ( a, b ) = f ( . . . f ( f ( f ( F, x ′ ) , B ) [ k ] , D ) . . . )where D = x . Writing f ( f ( F, x ′ ) , B ) as f ( . . . f ( f ( f ( F, x ′ ) , B ) [ k ] , x ′′′ ) . . . ) where x ′′′ := x , we see by Lemma 20 that deg(maxt( E )) ≤ deg(maxt( x ′′′ )) for every f -expression E in the ellipses of f ( . . . f ( f ( f ( F, x ′ ) , B ) [ k ] , x ′′′ ) . . . ). It follows by Lemma18 that e ( f ( a, b )) > L e ( f ( f ( F, x ′ ) , B )), which is a contradiction. Thus, we must have f ( a, b ) [ k +1] = f ( f ( f ( F, x ′ ) , B ) [ k ] , x ) = f ( f ( F, x ′ ) , B ) [ k +1] . This completes the induction.Consequently, we have f ( a, b ) [( m − n ] = f ( F, x ), so it follows that a = f ( F, x ). Finally, e ( f ( f ( F, x ) , b )) = e ( f ( f ( F, x ) , B )) implies that e ( b ) = e ( B ).Next, we will apply the concept of lexicographic minimality to prove another generalresult. First, let A , B be fixed f -expressions. We examine some of the restrictions e ( f ( A, B )) = e ( f ( C, D )) imposes on the f -expressions C and D . To avoid triviality,assume e ( B ) = e ( D ). We have e ( A ) + e ( B ) = e ( C ) + e ( D ) iff e ( B ) − e ( D ) = e ( C ) − e ( A ) (5)11ff ( e ( B ) − e ( D ))( e ( B ) + e ( B ) e ( D ) + e ( D ) ) = e ( C ) − e ( A ) . Since e ( B ) − e ( D ) =0, we know that e ( B ) − e ( D ) is not a constant. So we have deg( e ( C ) − e ( A ) ) > max(2 dege( B ) , D )). In particular, if dege( A ) ≤ dege( B ), then we must have2 dege( C ) > max(2 dege( B ) , D )) . (6)This implies that 2 dege( C ) > B ) ≥ A ), which implies that dege( C ) > dege( A ). It then follows by (5) that e ( B ) > L e ( D ) , so we have e ( B ) > L e ( D ). Ofcourse, from (6) we also obtain dege( C ) > dege( D ). Therefore, we have established thefollowing Lemma 28.
Let A , B be fixed f -expressions such that dege( A ) ≤ dege( B ) . If e ( f ( A, B )) = e ( f ( C, D )) and e ( B ) = e ( D ) , then dege( C ) > dege( A ) , e ( B ) > L e ( D ) ,and dege( C ) > dege( D ) .Notation . In what follows, we will denote the coefficient of the high-est degree term of a polynomial p ( x ) by lead( p ( x )). For example, we havelead( e ( f ( f ( x, f ( x, x )) , f ( f ( x, x ) , x )))) = 2.For the next two propositions, fix f ( A, f ( E, B ( m ) )) ∈ term( f ; x ) where dege( E ) ≤ dege( B ( m ) ) and dege( A ) ≤ dege( f ( E, B ( m ) )), so e ( f ( A, f ( E, B ( m ) ))) is monic, of degree3 m +2 . Writing f ( A, f ( E, B ( m ) )) as f ( A, f ( E, f ( x ′ , B ( m − ))), we see that 2 dege( A ) < deg(maxt( E )) < deg(maxt( x ′ )), because dgap( A, f ( E, B ( m ) )) = 3 m +2 − A ) ≥ m +2 − f ( E, B ( m ) )) = 3 m +2 − · m +1 = 3 m +1 > m +1 − E ) =dgap( E, B ( m ) ) and dgap( E, B ( m ) ) = 3 m +1 − E ) ≥ m +1 − B ( m ) ) =3 m +1 − · m = 3 m > m − x ′ , B ( m − ). It follows that B ( m ) is e -isolated withrespect to f ( A, f ( E, B ( m ) )) by Corollary 24. Suppose that e ( f ( A, f ( E, B ( m ) ))) = e ( f ( C, D )) . Then D = f ( F, B ( m ) ) for some F ∈ term( f ; x ), so e ( f ( A, f ( E, B ( m ) ))) = e ( f ( C, f ( F, B ( m ) ))) . (7)In the following two propositions and their corollaries, we will use Lemma 28 along withthe concept of lexicographic minimality to show that, for a large class of f -expressionsthat E may assume, the preceeding equality implies that F = E & e ( C ) = e ( A ) . (8) Proposition 30.
Let f ( A, f ( E, B ( m ) )) ∈ term( f ; x ) have the property that dege( E ) ≤ dege( B ( m ) ) and dege( A ) ≤ dege( f ( E, B ( m ) )) . Suppose that e ( f ( A, f ( E, B ( m ) ))) = e ( f ( C, D )) . Then D = f ( F, B ( m ) ) for some F ∈ term( f ; x ) . Furthermore, we musthave e ( E ) ≥ L e ( F ) and dege( E ) = dege( F ) .Proof. We have already proved the first part of the conclusion in the discussion leadingup to (7), so it remains to show that e ( E ) ≥ L e ( F ) and dege( E ) = dege( F ). If e ( F ) = e ( E ), then we are done. Suppose e ( F ) = e ( E ). Then e ( f ( F, B ( m ) )) = e ( f ( E, B ( m ) )).12y Lemma 28 we must have dege( C ) > dege( A ) and e ( f ( E, B ( m ) )) > L e ( f ( F, B ( m ) )),and it follows that e ( E ) > L e ( F ) . (9)Therefore, if dege( E ) = dege( F ), we must have dege( E ) > dege( F ); so let us assumedege( E ) > dege( F ) and obtain a contradiction. We have e ( A ) + ( e ( E ) + e ( B ( m ) ) ) = e ( C ) + ( e ( F ) + e ( B ( m ) ) ) . After cancelling out the common e ( B ( m ) ) from both sides, we see that the degree of theleft-hand side is deg( e ( E ) e ( B ( m ) ) ) and the highest-degree term on the right-hand sidemust be e ( C ) because deg( e ( F ) e ( B ( m ) ) ) < deg( e ( E ) e ( B ( m ) ) ). This means that2 dege( C ) = deg( e ( E ) e ( B ( m ) ) ) . (10)Then the coefficient of the highest degree term of the left-hand side must belead(3 e ( E ) e ( B ( m ) ) ) = 3 lead( e ( E )) lead( e ( B ( m ) )) = 3 lead( e ( E )) , and the coef-ficient of the highest degree term of the right-hand side must be lead( e ( C ) ) =lead( e ( C )) . We must have lead( e ( C )) = 3 lead( e ( E )) , from which it follows thatlead( e ( C )) = √ e ( E )) , (11)which is not even rational. This is a contradiction, so we must have dege( E ) = dege( F ). Corollary 31.
Under the hypotheses of Proposition 30, if E induces the lexicographi-cally minimal polynomial with degree dege( E ) , then F = E and e ( C ) = e ( A ) .Proof. We must have e ( E ) = e ( F ) or e ( E ) < L e ( F ). Since e ( E ) ≥ L e ( F ) by Proposition30, this forces e ( F ) = e ( E ), from which the conclusion follows by Lemma 26.The next proposition will be a generalization of Proposition 30. We will demonstratein Corollary 34 that (8) follows even if f ( E, B ( m ) ) assumes values in a more general classof f -expressions than that in Corollary 31. Notation . For every integer 1 ≤ j ≤ m define the function Y j : term( f ; x ) −→ term( f ; x ) by Y j ( A ) = f ( A, B ( j ) ) for every A ∈ term( f ; x ); in other words, we can write Y j = f ( − , B ( j ) ) as a one-argument function. Proposition 33.
Assume Notation 32 above. Let ≤ j ≤ m be a positive integerand d , . . . , d j − , d j be a sequence of positive integers such that m = d > d >. . . > d j − > d j ≥ . Let U ∈ term( f ; x ) with dege( U ) ≤ dege( B ( d j ) ) = 3 d j , andlet E = Y d ( Y d ( . . . Y d j ( U ) . . . )) , so f ( E, B ( m ) ) = Y d ( E ) = Y d ( Y d ( . . . Y d j ( U ) . . . )) = Y d ◦ Y d ◦ . . . ◦ Y d j ( U ) . Let A ∈ term( f ; x ) be such that dege( A ) ≤ dege( f ( E, B ( m ) )) .Suppose that e ( f ( C, D )) = e ( f ( A, f ( E, B ( m ) ))) . Then there exists V ∈ term( f ; x ) suchthat D = Y d ◦ Y d ◦ . . . ◦ Y d j ( V ) , e ( U ) ≥ L e ( V ) , and dege( V ) = dege( U ) . roof. We will prove this by induction on j . We have already proved the case j = 1in Proposition 30. Now assume that the statement of this proposition holds for all1 ≤ k ≤ j −
1. In what follows, we will prove this proposition for j .Since e ( f ( C, D )) = e ( f ( A, f ( E, B ( m ) ))) and f ( E, B ( m ) ) = Y d ◦ Y d ◦ . . . ◦ Y d j − ( f ( U, B ( d j ) )), it follows from our inductive hypothesis (with f ( U, B ( d j ) ) in therole of U ) that D = Y d ◦ Y d ◦ . . . ◦ Y d j − ( ˜ V ) for some ˜ V ∈ term( f ; x ) such that e ( f ( U, B ( d j ) )) ≥ L e ( ˜ V ) and dege( ˜ V ) = dege( f ( U, B ( d j ) )). We claim that ˜ V = f ( V, B ( d j ) ) for some V ∈ term( f ; x ) such that e ( V ) ≤ L e ( U ). Notice that we must have˜ V = f ( V, W ) for some
V, W ∈ term( f ; x ), where dege( W ) = dege( B ( d j ) ). Notice alsothat f ( U, B ( d j ) ) = f ( U, f ( x ′ , B ( d j − )) where x ′ := x , and deg( e ( U ) ) = 2 dege( U ) ≤ B ( d j ) ) < B ( d j ) ) = deg( x ′ )+2 deg( e ( B ( d j − ) ) = deg(maxt( x ′ )). Sup-pose W = B ( d j ) . Then by Lemma 25 we have e ( f ( U, B ( d j ) )) < L e ( f ( V, W )) = e ( ˜ V ),which is a contradiction. Hence we must have W = B ( d j ) . Since e ( f ( U, B ( d j ) )) ≥ L e ( ˜ V ) = e ( f ( V, B ( d j ) )), we have e ( V ) ≤ L e ( U ) as claimed.Now we want to show that dege( V ) = dege( U ). Supposedege( V ) < dege( U ) . (12)We will show that this leads to a contradiction. We have e ( f ( A, f ( E, B ( m ) ))) = e ( f ( A, Y d ◦ Y d ◦ . . . ◦ Y d j ( U ))) = e ( A ) +((( . . . (( e ( U ) + e ( B ( d j ) ) ) + e ( B ( d j − ) ) ) . . . ) + e ( B ( d ) ) ) + e ( B ( d ) ) ) = e ( C ) + ((( . . . (( e ( V ) + e ( B ( d j ) ) ) + e ( B ( d j − ) ) ) . . . ) + e ( B ( d ) ) ) + e ( B ( d ) ) ) = e ( f ( C, Y d ◦ Y d ◦ . . . ◦ Y d j ( V ))) = e ( f ( C, D )). Notice that,in the polynomial expansion of the preceeding five equalities, the terms (excludingthe e ( A ) and e ( C ) ) that do not contain e ( U ) as a factor or e ( V ) as a factor,( e ( B ( d ) ) ) ( e ( B ( d ) ) ) for example, are common to both sides of the third equalityand can thus be subtracted off from these two sides. Subtracting off these commonterms leaves deg( e ( U ) e ( B ( d j ) ) e ( B ( d j − ) ) . . . e ( B ( d ) ) e ( B ( d ) ) )as the degree of the left-hand side of the third equality. Sincedeg( e ( V ) e ( B ( d j ) ) e ( B ( d j − ) ) . . . e ( B ( d ) ) e ( B ( d ) ) )is less than the degree of the left-hand side of the third equality by (12), it fol-lows that 2 dege( C ) must equal the degree of the left-hand side of the third equal-ity. We see from the third equality that the coefficient of the highest degree termof the left-hand side is lead(3 · j − e ( U ) e ( B ( d j ) ) e ( B ( d j − ) ) . . . e ( B ( d ) ) e ( B ( d ) ) ) =3 · j − lead( e ( U )) lead( e ( B ( d ) )) = 3 · j − lead( e ( U )) and that the coefficient ofthe highest degree term of the right-hand side is lead( e ( C ) ) = lead( e ( C )) . Itfollows that we must have lead( e ( C )) = 3 · j − lead( e ( U )) , which implies thatlead( e ( C )) = √ · j − lead( e ( U )), which is not even rational. This is a contradiction,so we must have dege( U ) = dege( V ). This completes the induction. Corollary 34.
Under the hypotheses of Proposition 33, if e ( U ) is lexicographicallyminimal among polynomials with degree dege( U ) induced by f -expressions, then V = U and e ( A ) = e ( C ) . roof. We must have e ( V ) = e ( U ) or e ( V ) > L e ( U ). Since e ( U ) ≥ L e ( V ) by Proposition33, we must have e ( V ) = e ( U ), from which the conclusion follows by Lemma 26.In the proof of Proposition 30 (and analogously Proposition 33), we saw that, underthe hypotheses of this proposition and given dege( E ) > dege( F ), the subexpression C is not able to “make up” for the difference between e ( f ( E, B ( m ) )) and e ( f ( F, B ( m ) )),and hence e ( f ( C, f ( F, B ( m ) ))) = e ( f ( A, f ( E, B ( m ) ))). We will analyze and make use ofthis phenomenon extensively in what follows, where we consider the developments of f -expressions more complicated than A of Proposition 22.Consider A, B , E ∈ term( f ; x ) where B , E are subexpressions of A , dege( A ) =2 p q , dege( B ) = 2 m n , and dege( E ) = 2 i j . For the remainder of this paper, assumethe following three Assumption . B contains all the cores of A and is e -isolated with respect to A . Assumption . A [ m + n +1] is either f ( E , B ) or f ( B , E ). In other words, we can write A = f ( . . . f ( E , B ) . . . ) or A = f ( . . . f ( B , E ) . . . ) respectively. Assumption . For every occurrence x ′ of x in E and every occurrence x ′′ of x inthe ellipses of the expressions for A shown in Assumption 36 we have deg(maxt( x ′ )) > deg(maxt( x ′′ )). Remark . Note that Assumption 37 is a more general version of the correspondinghypothesis in Proposition 22.In the restricted version of Friedman’s problem we will study below, we will showthat whether or not A [ m + n +1] is e -isolated with respect to A is related to the solutionsets of certain exponential Diophantine equations.In either of the cases A = f ( . . . f ( E , B ) . . . ) or A = f ( . . . f ( B , E ) . . . ) we havedgap( E , B ) = 2 m + π n + π − i + π j + π (13)where { π , π } = { , } . Note that π = 1 corresponds to the case A = f ( . . . f ( B , E ) . . . ) and π = 1 corresponds to the case A = f ( . . . f ( E , B ) . . . ). Wecould attempt to prove, as in Proposition 22, that f ( E , B ) or f ( B , E ) is e -isolatedwith respect to A , but this assertion, if true, may be very difficult to prove. Below,we will simply explore how this problem can be analyzed through the study of certainDiophantine equations. Here are some lemmas that will aid us in this effort. Lemma 39.
Let A = f ( . . . f ( E , B ) . . . ) and A ′ = f ( . . . f ( E , B ) . . . ) be f -expressionssuch that dege( A ) = 2 p q = dege( A ′ ) , B contains all the cores of A , and B containsat least one core of A ′ . Suppose that for every occurrence x of x in E and for everyoccurrence x of x in the ellipses of A we have deg(maxt( x )) > deg(maxt( x )) . Supposethat e ( E ) < L e ( E ) . Then e ( A ) < L e ( A ′ ) .Proof. We can write e ( E ) = p ( x )+ q ( x ) and e ( E ) = p ( x )+ q ( x ), where p ( x ), p ( x ), q ( x ) are polynomials and deg( p ( x )) < deg( p ( x )). Examining ( e ( E ) + e ( B ) ) p q B ) from e ( A ) and ( e ( E ) + e ( B ) ) p q B ) from e ( A ′ ), we see that ( q ( x ) + e ( B ) ) p q B )
15s common to both e ( A ) and e ( A ′ ). We have deg( e ( A ) − ( q ( x ) + e ( B ) ) p q B ) ) =deg( p ( x )) + ( p q B ) − B )) < deg( p ( x )) + ( p q B ) − B )) ≤ deg( e ( A ′ ) − ( q ( x ) + e ( B ) ) p q B ) ), from which the conclusion follows.Analogously we have the following Lemma 40.
Let A = f ( . . . f ( B, E ) . . . ) and A ′ = f ( . . . f ( B, E ) . . . ) be f -expressionssuch that dege( A ) = 2 p q = dege( A ′ ) , B contains all the cores of A , and B containsat least one core of A ′ . Suppose that for every occurrence x of x in E and for everyoccurrence x of x in the ellipses of A we have deg(maxt( x )) > deg(maxt( x )) . Supposethat e ( E ) < L e ( E ) . Then e ( A ) < L e ( A ′ ) . Suppose ¯ A ∈ term( f ; x ) such that e ( ¯ A ) = e ( A ). Then we must have dege( ¯ A ) =dege( A ) = 2 p q and ¯ A [ m + n ] = B . Since B contains all the cores of A , B mustcontain all the cores of ¯ A ; otherwise we would have lead( e ( ¯ A )) > lead( e ( A )), whichis a contradiction. ¯ A [ m + n +1] must be either f ( E , B ) or f ( B , E ) for some E ∈ term( f ; x ), so we have either ¯ A = f ( . . . f ( E , B ) . . . ) or ¯ A = f ( . . . f ( B , E ) . . . ). Wewill say that A and ¯ A have the same orientation at the ( m + n + 1) st stage if A = f ( . . . f ( E , B ) . . . ) and ¯ A = f ( . . . f ( E , B ) . . . ), or if A = f ( . . . f ( B , E ) . . . )and ¯ A = f ( . . . f ( B , E ) . . . ). In each of the cases A = f ( . . . f ( E , B ) . . . ) or A = f ( . . . f ( B , E ) . . . ), our ultimate goal (which, by the way, will not be achieved in thispaper) is to show that A and ¯ A have the same orientation at the ( m + n + 1)st stageand furthermore that dgap( E , B ) = dgap( E , B ). In other words, in both cases wewant to show that dege( E ) = dege( E ), from which it would follow by Lemma 39 andLemma 40 that E = E whenever e ( E ) is lexicographically minimal for its degree.If dgap( E , B ) < dgap( E , B ), then deg(maxt( E )) = dege( A ) − dgap( E , B ) > dege( A ) − dgap( E , B ) = deg(maxt( E )), so under Assumption 37 we have e ( ¯ A ) > L e ( A ), which is a contradiction. Therefore, we must havedgap( E , B ) ≥ dgap( E , B ) . (14)Now, for the remainder of this paper, we make the following Assumption . Either A and ¯ A have the opposite orientation at the ( m + n + 1)st stageor dgap( E , B ) > dgap( E , B ).Notice that Assumption 41 is the negation of the statement “ A and ¯ A have thesame orientation at the ( m + n + 1)st stage and dgap( E , B ) = dgap( E , B )” whichwe hope to eventually prove, so we will try to derive contradictions under Assumption41. Also notice that Assumption 41 implies that A [ m + n +1] is not e -isolated with respectto A . We will see in the following discussion that certain exponential Diophantineequations must hold, and we will study these exponential Diophantine equations to seehow contradictions might be derived.There exists C ∈ term( f ; x ) and positive integer l ≥ m + n + 1 such that either¯ A [ l ] = f ( C, ¯ A [ l − ) or ¯ A [ l ] = f ( ¯ A [ l − , C ), and such that deg( e ( ¯ A ) − e ( B ) p q dege( B ) =16eg(maxt( C )). Since deg( e ( A ) − e ( B ) p q dege( B ) = deg(maxt( E )), we must also havedeg( e ( ¯ A ) − e ( B ) p q dege( B ) = deg(maxt( E )), so deg(maxt( C )) = deg(maxt( E )). Itfollows that dgap( C, ¯ A [ l − ) = dgap( E , B ). Let us now introduce subscripts that willshow more about the relation between C and B . Thus, there will exist (possibly morethan one choice of) k , k ∈ N ∪ { } and C k ,k ∈ term( f ; x ) (above called C ), such that¯ A [ m + n + k + k ] is either f ( C k ,k , ¯ A [ m + n + k + k − ) or f ( ¯ A [ m + n + k + k − , C k ,k ), k ≥ k ≥
1, dege( ¯ A [ m + n + k + k ] ) = 2 m + k n + k , (15)and dgap( C k ,k , ¯ A [ m + n + k + k − ) = dgap( E , B ); (16)note that, by our definition, k , k denote the number of times e ( B ) gets squared andthe number of times e ( B ) gets cubed, respectively, when we arrive at ¯ A [ m + n + k + k ] .We will call such a C k ,k a supplementing subexpression for E ; this is a gen-eralization of the “supplementing” C we mentioned in the paragraph following theproof of Corollary 34. Notice that the case k = π , k = π corresponds to thecase where A , ¯ A have the opposite orientation at the ( m + n + 1)st stage and wheredgap( E , B ) = dgap( E , B ); in this case we have C k ,k = C π ,π = E , i.e. E isone such supplementing subexpression for E . The case k = π , k = π correspondsto the case where A and ¯ A have the same orientation at the ( m + n + 1)st stage anddgap( C k ,k , ¯ A [ m + n + k + k − ) = dgap( E , B ). Since dgap( E , B ) > dgap( E , B ) byAssumption 41, we have dgap( C k ,k , ¯ A [ m + n + k + k − ) > dgap( E , B ), which is a con-tradiction. Therefore, we can exclude the case k = π , k = π (which we will callthe excluded case ) in our analysis below, because this case cannot happen underAssumption 41.Notice that either2 m + k n + k − dgap( C k ,k , ¯ A [ m + n + k + k − ) = 3dege( C k ,k )or 2 m + k n + k − dgap( C k ,k , ¯ A [ m + n + k + k − ) = 2dege( C k ,k ) , and both 3dege( C k ,k ), 2dege( C k ,k ) must be the product of a power of 2 and a powerof 3. Since 2 m + k n + k − dgap( C k ,k , ¯ A [ m + n + k + k − ) = 2 m + k n + k − dgap( E , B ) =2 m + k n + k − (2 m + π n + π − i + π j + π ), we must have2 m + k n + k − (2 m + π n + π − i + π j + π ) = 2 l l , (17)where l , l ∈ N ∪ { } . Solving (17) will make it easier for us to determine whether ornot the supplementing subexpression C k ,k exists and more generally whether or notAssumption 41 can be true. Notice that Equation (17) is essentially the equation2 a b + 2 c d = 2 e f + 2 g h . (18)We will study (18) in [1], where some partial results are proven.17 eferences [1] Roger Tian, On the Diophantine Equation a b + 2 c d = 2 e f + 2 g hh