On some diophantine equations in seperated variables
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Bérczes, Attila
A Diophantine equation has separated variables if it is of the form f(x)=g(y) for polynomials f, g. In a more general sense the degree of f and g may also be a variable.In the present talk various results for special types of the polynomials f and g will be presented. The types of the considered polynomials contain power sums, sums of products of consecutive integers, Komornik polynomials, perfect powers. Results on F-Diophantine sets, which are proved using results on Diophantine equations in separated variables will also be considered. The main tool for the proof of the presented general qualitative results is the famous Bilu-Tichy Theorem. Further, effective results (which depend on Baker’s method) and results containing the complete solutions to special cases of these equations will also be included.
 Thank you very much. Greetings to everyone. First of all, I would like to thank you very much the organizers to offer us this possibility. Of course, we all know that the best way to have a mass conference is to meet together in some place like Marseille. The second best thing is when this is not possible to have a virtual conference. So thank you very much for making the effort for this. So the plan of the talk is after a short introduction, I will say a few words about the BILUTI theorem and then I will address three topics in which with some co-authors I have a few results concerning the Ophantine equation in separated variables. Not all of them are connected to the BILUTI theorem. There are also some results which will be effective results. So first of all, if you take the Ophantine equation of the shape fxy equals zero, where fxy is a polynomial with rational coefficients, then in principle, the main question may be how this or when this equation may have a finite or infinitely many solutions. And of course, we know that if fxy is irreducible over q, but it is not absolutely irreducible, so it's not irreducible, let's say over c, then fxy equals zero has finitely many rational solutions. Denoting by G the genus and by D the number of points at infinity of the plane curve fxy equals zero, in 1929 Z-girl answered this question quite completely. In principle, his result says that if fxy is absolutely irreducible and G is larger or equal to one or D is larger equal to three, then one has only finitely many solutions. So Z-girl's result is very helpful when the polynomials are given explicitly or when they are of a very special form, but in principle it was for a long time a question whether we can describe all the pairs of polynomials f and G for which fxy equals G y, which is a special case of the previous shown equation, the case when the variables are separated in the left hand side y is on the right hand side, can we describe all the situations when this may have infinitely many solutions. In the type of equations with separated variables enter many important special cases, many important classes of the of anti-equations like super elliptic equations, hyper elliptic equations, Schindsel-Tiedemann equation, power values in arithmetic progressions or power values of the sum of case power, equal values of lecunary polynomials and several other classical and important the of anti-questions. Now the question of describing all pairs of polynomials such that the equation fx equals G y has infinitely many solutions, has been solved by Billow and Pih, in principle this is a widely known and appreciated theorem which became a very important tool in the of anti-number theory. They proved that if f and G are polynomials with rational coefficients which are non-constant, then the following two statements are equivalent. Equation 2 has infinitely many solutions in rational numbers with bounded denominator, in principle it's a kind of generalization of the of the searching for integer solutions. So this statement is equivalent to the fact that f is composed of three functions, three polynomials and G is composed again of three polynomials where the inner polynomials are linear polynomials, the middle polynomials form a so-called standard pair which on the next slide I will show how they look like and phi is just a polynomial with rational coefficients and it also has to be assumed that the equation fx equals G y does have infinitely many solutions in rational sweet bounded denominator. So in principle this is a very concise description of all the cases in which it is possible that this equation does have infinitely many solutions and there are the possible pairs for G and f or f and G of course in perpetuating the order doesn't make any difference. So there are five types of polynomials. The first type of polynomials look like x to the power q and in this product afterwards the second type is x squared and alpha x squared plus beta times vx squared where here alpha beta are nonzero rational numbers and the third types of and fourth type of pairs are defined using the big sum polynomials and finally there is a fifth pair where there are not many restrictions of the parameters. So these are the possibilities. Since this is a very concise and precise description of all the situations and there might be infinitely many solutions the ability to CRM became a widely used tool for proving finiteness of number of solutions in the case of the ofantian equations. I cannot list all or nearly all or not even the all the important results where the ability CRM has been used but I just show a few instances of the use of them maybe some of which I knew without searching for them. So Peter Pinter and Schindsel and later Schindsel alone applied the theory for separable equations in trinomials. Stoll and he investigated the problem for general makes and the crutch to polynomials. Bill Brinsock in Scherhofer Pinter and he used it for Bernoulli polynomials, Kresho for lacunary polynomials, Dubis Ketten Kresho for truncated binomial powers, Rokowski for polynomial values of power sums and Kulkarnian for polynomial values of products of consecutive numbers. So to start to present some results first of all let us define the sum of products of consecutive integers in principle this is a polynomial which is defined by the sum of the product of terms like x plus j so if we start to look at the first such polynomial f0 of x that's nothing else just x. f1 of x is x plus x times x plus 1. f2 of x will be x plus x times x plus 1 plus x times x plus 1 times x plus 2 and so on. So in principle what is clear from the definition is nothing else that f of x is a monic polynomial. Then it has positive integer coefficients. Its degree is k plus 1 and well as we see it is also visible from this that 0 and minus 2 will be always the root of this polynomial. These polynomials first were introduced as far as I know by Kojula, Estroman, Tenghe who proved that the the of Fanta, who considered the the of Fanta equation fk of x equals y to the n and they proved effective finiteness result for that. They even solved completely the equation when one smaller or equal to k smaller or equal to n and to do so they needed a few more involved properties of these polynomials. So about the fk of x here are the first few polynomials and if k is larger or equal to 3 then all the roots of the polynomial fk of x are real and simple. So the only case when there is a double root is here in f2 of x. 0 is always a root and minus 2 is always a root of the polynomial and a more precise description of the root structure of fk of x is here. These are the intervals where fk of x does have a root. So in principle each root can be can be put in an interval of at length most well 1.5 but generally 1. So it is very well described where are the roots of this polynomial. Okay so for our generalization of the result of Hore-Gula instrument Tangae we needed some properties of the polynomials fk of x plus 1 and f derivative of fk derivative of x sorry from here the case missing. So if k is larger or equal to 3 then all the roots of fk of x plus 1 are real and simple. That's one property which we will use later. Minus 1 is always a root of the polynomial fk of x plus 1 for all k larger or equal to 2. Again for k larger or equal to 2 we can describe where the roots of this polynomial fk x plus 1 beside. One of them is in the interval minus k minus 1 minus k the next one in minus k minus k plus 1 and so on. At 2 minus 3 minus 2 and finally there is one root in the interval minus 0.50. So for k larger or equal to 3 if we consider the derivative of such a polynomial then we can denote the roots by alpha 1 alpha 2 and so on. All these roots are real and simple so it will have exactly k roots. If k is larger or equal to 6 then we can prove that fk derivative is 0 is positive, fk derivative in the value minus 1 gives a negative value and so on. And for k larger or equal to 13 we just have fk alpha 1 is larger than fk alpha 3 is larger than fk alpha 5 and fk alpha 2 is larger than fk alpha 4. These are properties which we needed in the proofs of our results. Among the first results I will mention that fk of x is in the composable. So let us see what we mean by a polynomial being in the composable. By the composition of a polynomial f of x over a field k we mean that it is written as a composite function of two polynomial functions. So it is of the form g1 of g2 of x where g1 and g2 are both polynomials over the same field. And we say that such a decomposition is non-trivial if the degree of both these polynomial g1 and g2 is larger than 1. So if none of the polynomials g1 and g2 is linear. If we have two decompositions we say that these decompositions are equivalent. If there exists a linear polynomial with coefficients in the field k such that g1 can be expressed by h1 in the variable Lx and h2 of x is L of g2 of x. We will say that the polynomial f of x is decomposable over k if it has at least one non-trivial decomposition otherwise it is called indecomposable. And what we proved together with Vashu, Haidler and Luka was that the polynomial fk of x is indecomposable over the complex field for any k larger or equal to zero. To prove this we had two very nice tools, two lemmas of the yellow line usage. They proved that if fx is a polynomial with integer coefficients which is monic and it is decomposable over the complex field then there exists also decomposition over z. Further they also proved that if we have a polynomial f of x which is monic and has integer coefficients then in the case of any decomposition f of x equals gh of x the outer polynomial the degree of the outer polynomial is such that its degree is smaller or equal to the gcd of the second coefficient and the degree of the polynomial. So whenever we have a decomposition the outer polynomial is of degree less than the gcd of this coefficient and n the degree of the polynomial. This means that the gcd of a and n if this is co-prime then m is smaller or equal to one that means that if a and n are co-prime then any decomposition is with degree of g equals one which means that f is in decomposable so any decomposition would be a trivial decomposition. So let us see how can we use this for proving that fk of x is in decomposable. First of all let us see how fk of x looks like. We know that fk of x has degree k plus one it is monic so we just denote its coefficient by one ck minus one and so on c1 and we know that zero is a root of fk of x so c0 must be zero. It is not very hard I would say it is very easy to compute ck which is one plus k times k plus one over two. Now if you take a look at it this when k is even then this is k plus one times an integer if I add one that is co-prime to k plus one so in this case by the CRM of Duiola and Goussic fk of x is in decomposable this is the easy case when k is odd then we have to do a little bit more computations in this case we see that k plus one and ck has a gcd which is at most two. Well of course in the case when the degree of the outer polynomial if fk of x equals h of g of x if the degree of the outer polynomial is one that's a trivial decomposition so it doesn't count so the only way we may have a non-trivial decomposition is the degree of h is two because the degree of the outer polynomial the external polynomial must be smaller or equal to two. So assume that we have a quadratic polynomial h equals ax square plus bx plus c. Now then h of g of x looks like this and clearly now this is the decomposition of the polynomial fk of x but by the CRM of Duiola and Goussic we know that if it has a decomposition over the complex numbers it has a decomposition over the integers so we may assume without loss of generality that abc are integers and g is a polynomial with integer coefficients. From this it is easy to see that a must be equal to one because f has leading coefficient one so this means that a times the square of the leading coefficient of g must be one. All are integers so a must be one. This means that if you and we are the polynomial are the roots of the polynomial h the quadratic polynomial ax square or now x square plus bx plus c then fk of x can be written like g of x minus u times g of x minus v but we know that fk of zero is zero so either one or the other factor must be zero so we get g zero equals u or g zero equals v which now shows that either u or v is integer but now we have a quadratic equation if one of the roots is integer then the other one must be integers so both you and we are integers. So we may just denote gx minus u by h and then we know that gx minus v will be h of x plus d where d is u minus v but what is important then this d is an integer. So now we have fk of x equals h times h plus d and let us use the property of fk of x that fk of minus one is minus one that means that h minus one times h of minus one plus d they give minus one so since h of x and h of x plus d are integers then h of minus one equals plus or minus one and h of minus one plus d equals plus or minus one consequently we obtain d equals zero or d equals plus or minus two but d equals zero would mean that fk of x is h of x to d square but this contradicts that all the roots of f of x fk of x are simple roots and when we have d equals plus or minus two then we get that fk of x plus one will be a square but again we have proved that fk of x plus one has only simple roots which gives again a contradiction so all together fk of x is incomposable. This CRM was used to prove the second part of the following more general result let g of x be a polynomial with rational coefficients and consider the the of anti-inequation fk of x equals g of y so now the right hand side is an arbitrary polynomial and on the left hand side we do have this polynomial which is the sum of products of consecutive integers. Assume that k is larger or equal to three then in the case when the degree of the polynomial g is zero or two then there exists an effectively computable constant depending only on this index k and the polynomial g such that the maximum absolute value of the integer solutions of equation four is bounded by this constant so we can prove an effective finiteness CRM in this case whenever the degree of g is larger or equal to three then we can prove that equation four has only finitely many integer solution unless we have g of x equals fk h of x clearly in this case there will be infinitely many solutions so in the case of g equals zero or two we have effective finiteness and in the case of g larger or equal to three we do have ineffective finiteness here in the second point the finiteness is ineffective because we are using the bilou tihi criterion the bilou tihi CRM and since the bilou tihi CRM is ineffective the results proved by it will also be effective result this means that there are no bounds on the size of the solutions it proves a finiteness where there is no guaranteed that there is a bound we cannot just produce an effective upper bound well what about the cases which are excluded first of all whenever g of x equals fk of h of x then clearly we may have infinitely many solutions also when the degree of g equals one we may have infinitely many solutions well it is clear whenever g is linear then this might have infinitely many solutions let's say if g of y equals y then we just have fk of x equals an integer and if y is that integer that's a solution so for every x we get a solution y again when g of x is fk of h of x the same situation holds whenever h of x assumes a value then we can take for let's say for y or for the other unknown this value and we get a solution so in this case is we always have infinitely many solutions and it may happen also for k smaller or equal to two at the beginning of the CRM we assume k is larger or equal to three well when k is smaller or equal to two it may happen that there are infinitely many solutions these conditions are necessary now in case of our CRM now a few words about the proof of statement two I will not tell anything about the proof of the first statement but about the second statement where we use the ability CRM I would like to sketch the main ideas of the proof so it is this part of the CRM so as in the CRM we assume that k is larger or equal to three and the g is a polynomial with rational coefficients of degree at least three and we assume that this equation has infinitely many solutions now let us see what can be inferred from that the ability CRM describes very explicitly what situation this can happen fk of x must be phi of f of lambda of x and g must be phi of g of k of x the phi here is the same polynomial lambda and kappa they are linear polynomials and f and g are coming from a standard pair over q now we know we have proved a few slides before that fk of x is indecomposable so that means that the degree of this phi is either one or it is k plus one if it's one then of course f must be of degree k plus one otherwise f will be linear so let us see the two cases we will address first the case when degree of phi of x equals k plus one we just recall that this is how fk and g of x are composed now from this composition we see that the degree of f must be one because this is of degree k plus one phi is of degree k plus one so f must be linear but then clearly we have fk of x equals phi t of x where t of x is f lambda of x it's linear now we can take the inverse function of the polynomial function t of x so that t of t minus one gives the identity function so this means that fk of t minus x is just phi of x so we can express phi of x in the form fk of something which is linear now if we go to the other part of the or other polynomial g of x g of x is composed like phi of g of tappa of x but phi can be written like fk of t minus one and now you see that this is fk of something and in principle we don't care what is here inside the only important feature is that it is a polynomial with rational coefficients and of degree at least one so this means that we can prove that g of x equals fk of h of x this is the exceptional case when we may have infinitely many solutions and we will see that there are no other possibilities for infinitely many solutions since the other possibility is when the degree of phi of x equals one whenever the degree of phi of x equals one we can write phi of x like this and in this case we have to study all the possible standard pairs now we have five kinds of standard pairs in the case of standard pairs of type two this can be excluded because there we have a polynomial of degree two but f of x and g of x cannot be of degree two in this case because f of x must be of degree k plus one if i go back here now phi is linear lambda is linear so in order to get the polynomial of degree k plus one f must be of degree k plus one it cannot be quadratic polynomial so on the other hand the other polynomial let me just come back for a second here to the second type standard pairs so one of them must be x square and the other one must be of this shape but here it's clear that this polynomial has multiple roots whenever the degree is larger than two or larger than larger equal to four sorry so in that case again we see that it has multiple roots but f k of x cannot have multiple roots so this way we can exclude polynomials of pairs of polynomials of type two standard pairs of type two standard pairs of certain force kind they are the polynomials which are defined using with some polynomials we can exclude them by comparing coefficients in this equation in the case of standard pairs of type five in principle we have two cases f k of x must be either of this shape which contradicts with the in the composibility of f if we look at this then this shows that there is a polynomial eight times lambda x square minus one to the cube but this means that this polynomial is composed to phi x cube plus phi one x cube plus phi zero so this is a non-trivial composition of polynomials f k cannot be of this shape on the other hand we may have f k of x equals this shape and in this case we just compare again coefficients and after comparing three four coefficients we get a contradiction in the case of the first type again the root structure of f k of x and the derivative of f k of x prove that such standard pairs are impossible well of course here we have some lengthier computation which I have no time to to show for these situations but this is the flavor of the of the proof we also proved an effective result in the case of the equation f k of x equals a y to the n plus b we could prove that the solutions are bounded by an effectively computable constant and well this is a generalization of the result of Hoidl-Lichram and Tange who considered the case a equals one and b equals zero now we will switch our attention to a different topic. Comornic polynomials are polynomials introduced by Riemann-Kommornig for some results concerning digital expansions and if we define a sub n of x equals x to the power two times three to the power n minus x to the power three to the n minus one and we multiply such polynomials from k equals zero to n then we get the n's Komornig polynomial. Komornig polynomials the first few of them look like this the first one is easy to see but the second one and the third one are these polynomials so we can see that these are polynomials with coefficients plus or minus one and there are the degree of this of such polynomials is always c to the power n plus one minus one. All coefficients are plus or minus one below the highest degree term below the main term there are no other coefficients than plus or minus one including zero there is no zero coefficient so all the terms are present it consists of c to the power n plus one monomials and all these monomials have coefficients either plus one or minus one. It is also proved that all the complex roots of such polynomials are simple. The first important result concerning Komornig polynomials is that Komornig polynomials are incomposable over c and this helps us to use this for to apply the B-Lutih C-R-M later but for the moment let us see a few more properties of a k of x. In principle there is a polynomial identity a l plus k of x equals a of a sub l of x to the power three to the power k and if we denote the Fibonacci numbers but by phi zero, pi one and so on and we define f sub k to be phi of three to the power k plus one plus phi of three to the k minus one minus one then these are clearly integer numbers these integer numbers will play an important role in the GCD of two values of this polynomial a k of x. So if we take the polynomial a k of x in any integer space and we compute it for the index l and for the index l plus k then the GCD of these two integers will divide the integer a sub k defined just here using the Fibonacci numbers. So this way for polynomials with index close to each other we can compute the GCD or we can prove how much is the GCD if they are consecutive in this is k and k plus one then the GCD will always be one and if between the indexes between the indices the difference is two then the GCD is either one or five. So after these auxiliary properties let us show about show the results concerning the of anti-equations containing Pomornik polynomials the problem originates from Atilla Petu who asked in a conference about the finiteness of number of solutions of e sub n of x equals e sub n of y. With Kolushpink and Schustek we did prove that the equation p sub n of x equals p sub m of y has always finitely many integer solutions and even more general we consider the equation with arbitrary right hand side and we did prove that if g of x is not of the form p sub n of h of x or g of x is not of the form gamma of delta x to the l where gamma and delta are linear polynomials directional coefficients and l is just the degree of this polynomial g then we have only finitely many solutions. However I have to mention that the first exception that's necessary the second exception we couldn't prove that in this case there are infinitely many solutions so it may it probably may be excluded but we were just unable to prove that. Well that was what it was written here and of course since we use BDT-CRM the result is ineffective. We also did prove some results concerning this polynomial which are effective in the case of the equation p of n of x equals y to the m. We did get effective solutions both in the case m equals to or m larger would be equal to 3 and we even could prove an effective result but for the exponent m from here whenever y is different of plus or minus one and for the concrete cases of y square in the right hand side we did solve completely this equation there are some trivial solutions and except of the trivial solutions it exists only one non-trivial solution in the case when n equals 2 and we did solve the equation also for n equals 1, 2, 3, 4 and 5. Further we did prove we did solve completely the equation p of n of x equals y cube for n equals 0, 1, 2 and 3. As my time is going I will not tell about these results in details however I will switch to my third topic power values of power sums. Well if you take s k of x 1 to the k plus 2 to the k plus x to the k this is s k of x and the problem s k of x equals y to the n this is a well-known and much-investigated classical problem it goes back to Luca Watson and the first international known base through was you to check for could prove that if k and n is not one of these pairs then this equation 16 has only 590 many solutions. He also conjectured that if k and n are not k and is not of this pair not not one of these pairs then the only non-trivial solution is given by 2, 2, 24 and 70. Many results concerning this equation have been established during the last few decades. Yuri Tiedemann and Forkhoof proved an effective version of Schaeffer and they have another extra feature that also n was an unknown so in this equation also the exponent was an unknown for them and the effective finiteness was working also in this more general case. Then Pinter proved that for non-trivial solutions we have this bound and the conjecture was verified by Jacobson Pinter and Walsh for some bounded values of k and n equals 2. Benadio and Pinter extended the result to one smaller or equal to k smaller or equal to 11 and arbitrary and larger or equal to 2. This was again an important breakthrough and Pinter verified the conjecture for and larger than 4 and one smaller or equal to k smaller than 107. This is of course not a complete list of results. However, how do we initiate the investigation of this equation for fixed values of x? For values of x which are smaller than 25 and are congruent of 0 and 3 mod 4, he computed all the solutions and together with Koidu Mierzak and Inc. we extended this result for the cases when k is congruent to one or two mod 4. In principle this shows that we proved the Schaeffer conjecture for x smaller than 25 and larger or equal to 3. Now finally I would like to mention few other kinds of equations where S k of x is included and the Bielow-Tihe CRM is used, but these results are not mine. I couldn't contribute to this topic. Bielow-Grinsow-Kinscher Hofer-Pieter and Tihe consider the equation S k of x equals SM of y and for the case when Geofi is the product of consecutive integers Bielow-Binser Hofer-Pieter and Tihe proved again, then Jürj Kovac, Peter and Inter for binomial coefficients and for this kind of sums Jürj Kovac, Peter and Inter. So also for such kind of separable equations Bielow-Tihe CRM was used and it was a powerful tool. Thank you very much for your attention. This was what I wanted to tell. Yeah so thank you very much Attila for your very nice and rich talk. So it's open for discussion and I already can see a very precise question by Gary Walsh who is our next speaker. So the question you can read here is it possible to read it? Just a second I'm trying to find out. So he is asking where do the upper pounds come on page 28? On page 28. Very precise question. Page 28 just let me find oh no I'm sorry no no it's 20 oh yeah 28. I see well these are coming from super and hyper elliptic equations from Baker's theory from a result of Jahn de Kewer, Czekalman, Jürj and myself. We have a an explicit a completely explicit result for this case and we use that CRM for this.
 Thank you very much. Greetings to everyone. First of all, I would like to thank you very  much the organizers to offer us this possibility. Of course, we all know that the best way to  have a mass conference is to meet together in some place like Marseille. The second best  thing is when this is not possible to have a virtual conference. So thank you very much  for making the effort for this. So the plan of the talk is after a short introduction,  I will say a few words about the BILUTI theorem and then I will address three topics in which  with some co-authors I have a few results concerning the Ophantine equation in separated  variables. Not all of them are connected to the BILUTI theorem. There are also some results  which will be effective results. So first of all, if you take the Ophantine equation of the shape  fxy equals zero, where fxy is a polynomial with rational coefficients, then in principle,  the main question may be how this or when this equation may have a finite or infinitely many  solutions. And of course, we know that if fxy is irreducible over q, but it is not absolutely  irreducible, so it's not irreducible, let's say over c, then fxy equals zero has finitely many  rational solutions. Denoting by G the genus and by D the number of points at infinity of the  plane curve fxy equals zero, in 1929 Z-girl answered this question quite completely. In principle,  his result says that if fxy is absolutely irreducible and G is larger or equal to one or D is larger  equal to three, then one has only finitely many solutions. So Z-girl's result is very helpful  when the polynomials are given explicitly or when they are of a very special form, but in  principle it was for a long time a question whether we can describe all the pairs of polynomials f and  G for which fxy equals G y, which is a special case of the previous shown equation, the case  when the variables are separated in the left hand side y is on the right hand side, can we  describe all the situations when this may have infinitely many solutions. In the type of equations  with separated variables enter many important special cases, many important classes of the  of anti-equations like super elliptic equations, hyper elliptic equations, Schindsel-Tiedemann  equation, power values in arithmetic progressions or power values of the sum of case power,  equal values of lecunary polynomials and several other classical and important the of anti-questions.  Now the question of describing all pairs of polynomials such that the equation fx equals G y  has infinitely many solutions, has been solved by Billow and Pih, in principle this is a widely  known and appreciated theorem which became a very important tool in the of anti-number theory.  They proved that if f and G are polynomials with rational coefficients which are non-constant,  then the following two statements are equivalent. Equation 2 has infinitely many solutions in  rational numbers with bounded denominator, in principle it's a kind of generalization of the  of the searching for integer solutions. So this statement is equivalent to the fact that f is  composed of three functions, three polynomials and G is composed again of three polynomials where  the inner polynomials are linear polynomials, the middle polynomials form a so-called standard pair  which on the next slide I will show how they look like and phi is just a polynomial with  rational coefficients and it also has to be assumed that the equation fx equals G y does have  infinitely many solutions in rational sweet bounded denominator. So in principle this is a very  concise description of all the cases in which it is possible that this equation does have  infinitely many solutions and there are the possible pairs for G and f or f and G of course  in perpetuating the order doesn't make any difference. So there are five types of polynomials.  The first type of polynomials look like x to the power q and in this product afterwards the second  type is x squared and alpha x squared plus beta times vx squared where here alpha beta are nonzero  rational numbers and the third types of and fourth type of pairs are defined using the  big sum polynomials and finally there is a fifth pair where there are not many restrictions of the  parameters. So these are the possibilities. Since this is a very concise and precise description  of all the situations and there might be infinitely many solutions the ability to CRM became a widely  used tool for proving finiteness of number of solutions in the case of the ofantian equations.  I cannot list all or nearly all or not even the all the important results where the ability  CRM has been used but I just show a few instances of the use of them maybe some of which I knew  without searching for them. So Peter Pinter and Schindsel and later Schindsel alone applied the  theory for separable equations in trinomials. Stoll and he investigated the problem for  general makes and the crutch to polynomials. Bill Brinsock in Scherhofer Pinter and he used it  for Bernoulli polynomials, Kresho for lacunary polynomials, Dubis Ketten Kresho for truncated  binomial powers, Rokowski for polynomial values of power sums and Kulkarnian for polynomial  values of products of consecutive numbers. So to start to present some results first of all  let us define the sum of products of consecutive integers in principle this is a polynomial  which is defined by the sum of the product of terms like x plus j so if we start to look at  the first such polynomial f0 of x that's nothing else just x. f1 of x is x plus x times x plus 1.  f2 of x will be x plus x times x plus 1 plus x times x plus 1 times x plus 2 and so on.  So in principle what is clear from the definition is nothing else that f of x is a monic polynomial.  Then it has positive integer coefficients. Its degree is k plus 1 and well as we see it is also  visible from this that 0 and minus 2 will be always the root of this polynomial.  These polynomials first were introduced as far as I know by Kojula, Estroman,  Tenghe who proved that the the of Fanta, who considered the the of Fanta equation fk of  x equals y to the n and they proved effective finiteness result for that. They even solved  completely the equation when one smaller or equal to k smaller or equal to n and to do so they needed  a few more involved properties of these polynomials. So about the fk of x here are the first few  polynomials and if k is larger or equal to 3 then all the roots of the polynomial fk of x are real  and simple. So the only case when there is a double root is here in f2 of x. 0 is always a root and  minus 2 is always a root of the polynomial and a more precise description of the root structure  of fk of x is here. These are the intervals where fk of x does have a root. So in principle  each root can be can be put in an interval of at length most well 1.5 but generally 1. So it is  very well described where are the roots of this polynomial. Okay so for our generalization of  the result of Hore-Gula instrument Tangae we needed some properties of the polynomials fk of x plus  1 and f derivative of fk derivative of x sorry from here the case missing. So if k is larger or  equal to 3 then all the roots of fk of x plus 1 are real and simple. That's one property which we  will use later. Minus 1 is always a root of the polynomial fk of x plus 1 for all k larger or  equal to 2. Again for k larger or equal to 2 we can describe where the roots of this polynomial fk x  plus 1 beside. One of them is in the interval minus k minus 1 minus k the next one in minus k minus  k plus 1 and so on. At 2 minus 3 minus 2 and finally there is one root in the interval minus 0.50.  So for k larger or equal to 3 if we consider the derivative of such a polynomial then we can denote  the roots by alpha 1 alpha 2 and so on. All these roots are real and simple so it will have exactly  k roots. If k is larger or equal to 6 then we can prove that fk derivative is 0 is positive,  fk derivative in the value minus 1 gives a negative value and so on. And for k larger or  equal to 13 we just have fk alpha 1 is larger than fk alpha 3 is larger than fk alpha 5 and fk  alpha 2 is larger than fk alpha 4. These are properties which we needed in the proofs of our  results. Among the first results I will mention that fk of x is in the composable. So let us see  what we mean by a polynomial being in the composable. By the composition of a polynomial f of x over  a field k we mean that it is written as a composite function of two polynomial functions. So it is of  the form g1 of g2 of x where g1 and g2 are both polynomials over the same field. And we say that  such a decomposition is non-trivial if the degree of both these polynomial g1 and g2 is larger than  1. So if none of the polynomials g1 and g2 is linear. If we have two decompositions we say that  these decompositions are equivalent. If there exists a linear polynomial with coefficients in the  field k such that g1 can be expressed by h1 in the variable Lx and h2 of x is L of g2 of x.  We will say that the polynomial f of x is decomposable over k if it has at least one  non-trivial decomposition otherwise it is called indecomposable. And what we proved together with  Vashu, Haidler and Luka was that the polynomial fk of x is indecomposable over the complex field  for any k larger or equal to zero. To prove this we had two very nice tools, two lemmas of the  yellow line usage. They proved that if fx is a polynomial with integer coefficients which is  monic and it is decomposable over the complex field then there exists also decomposition over z.  Further they also proved that if we have a polynomial f of x which is monic and has  integer coefficients then in the case of any decomposition f of x equals gh of x the outer  polynomial the degree of the outer polynomial is such that its degree is smaller or equal to the  gcd of the second coefficient and the degree of the polynomial. So whenever we have a decomposition  the outer polynomial is of degree less than the gcd of this coefficient and n the degree of the  polynomial. This means that the gcd of a and n if this is co-prime then m is smaller or equal to  one that means that if a and n are co-prime then any decomposition is with degree of g equals one  which means that f is in decomposable so any decomposition would be a trivial decomposition.  So let us see how can we use this for proving that fk of x is in decomposable. First of all let  us see how fk of x looks like. We know that fk of x has degree k plus one it is monic so we just  denote its coefficient by one ck minus one and so on c1 and we know that zero is a root of fk of  x so c0 must be zero. It is not very hard I would say it is very easy to compute ck which is one  plus k times k plus one over two. Now if you take a look at it this when k is even then this is k  plus one times an integer if I add one that is co-prime to k plus one so in this case by the  CRM of Duiola and Goussic fk of x is in decomposable this is the easy case when k is odd then we have  to do a little bit more computations in this case we see that k plus one and ck has a gcd which is  at most two. Well of course in the case when the degree of the outer polynomial if fk of x equals  h of g of x if the degree of the outer polynomial is one that's a trivial decomposition so it doesn't  count so the only way we may have a non-trivial decomposition is the degree of h is two because  the degree of the outer polynomial the external polynomial must be smaller or equal to two.  So assume that we have a quadratic polynomial h equals ax square plus bx plus c. Now then h of g of x  looks like this and clearly now this is the decomposition of the polynomial fk of x but by  the CRM of Duiola and Goussic we know that if it has a decomposition over the complex numbers it has  a decomposition over the integers so we may assume without loss of generality that abc are integers  and g is a polynomial with integer coefficients. From this it is easy to see that a must be equal to one  because f has leading coefficient one so this means that a times the square of the leading  coefficient of g must be one. All are integers so a must be one. This means that if you and we are  the polynomial are the roots of the polynomial h the quadratic polynomial ax square or now x square  plus bx plus c then fk of x can be written like g of x minus u times g of x minus v but we know that  fk of zero is zero so either one or the other factor must be zero so we get g zero equals u  or g zero equals v which now shows that either u or v is integer but now we have a quadratic equation  if one of the roots is integer then the other one must be integers so both you and we are integers.  So we may just denote gx minus u by h and then we know that gx minus v will be h of x plus d where  d is u minus v but what is important then this d is an integer.  So now we have fk of x equals h times h plus d and let us use the property of fk of x that fk of  minus one is minus one that means that h minus one times h of minus one plus d they give minus one  so since h of x and h of x plus d are integers then h of minus one equals plus or minus one  and h of minus one plus d equals plus or minus one consequently we obtain d equals zero or d  equals plus or minus two but d equals zero would mean that fk of x is h of x to d square  but this contradicts that all the roots of f of x fk of x are simple roots and when we have d equals  plus or minus two then we get that fk of x plus one will be a square but again we have proved that  fk of x plus one has only simple roots which gives again a contradiction so all together  fk of x is incomposable. This CRM was used to prove the second part of the following more  general result let g of x be a polynomial with rational coefficients and consider the  the of anti-inequation fk of x equals g of y so now the right hand side is an arbitrary polynomial  and on the left hand side we do have this polynomial which is the sum of products of  consecutive integers. Assume that k is larger or equal to three then in the case when the degree of  the polynomial g is zero or two then there exists an effectively computable constant  depending only on this index k and the polynomial g such that the maximum absolute value of the  integer solutions of equation four is bounded by this constant so we can prove an effective  finiteness CRM in this case whenever the degree of g is larger or equal to three then we can prove  that equation four has only finitely many integer solution unless we have g of x equals fk h of x  clearly in this case there will be infinitely many solutions  so in the case of g equals zero or two we have effective finiteness and in the case of g larger  or equal to three we do have ineffective finiteness here in the second point the finiteness is  ineffective because we are using the bilou tihi criterion the bilou tihi CRM and since the bilou  tihi CRM is ineffective the results proved by it will also be effective result this means that  there are no bounds on the size of the solutions it proves a finiteness where there is no guaranteed  that there is a bound we cannot just produce an effective upper bound well what about the cases  which are excluded first of all whenever g of x equals fk of h of x then clearly we may have  infinitely many solutions also when the degree of g equals one we may have infinitely many  solutions well it is clear whenever g is linear then this might have infinitely many solutions  let's say if g of y equals y then we just have fk of x equals an integer and if y is that integer  that's a solution so for every x we get a solution y again when g of x is fk of h of x  the same situation holds whenever h of x assumes a value then we can take for let's say for y  or for the other unknown this value and we get a solution so in this case is we always have  infinitely many solutions and it may happen also for k smaller or equal to two at the beginning  of the CRM we assume k is larger or equal to three well when k is smaller or equal to two  it may happen that there are infinitely many solutions  these conditions are necessary now in case of our CRM now a few words about the proof of  statement two I will not tell anything about the proof of the first statement but about the  second statement where we use the ability CRM I would like to sketch the main ideas of the proof  so it is this part of the CRM  so as in the CRM we assume that k is larger or equal to three and the g is a polynomial with  rational coefficients of degree at least three and we assume that this equation has infinitely many  solutions now let us see what can be inferred from that the ability CRM describes very explicitly  what situation this can happen fk of x must be phi of f of lambda of x and g must be phi of g  of k of x the phi here is the same polynomial lambda and kappa they are linear polynomials  and f and g are coming from a standard pair over q  now we know we have proved a few slides before that fk of x is indecomposable  so that means that the degree of this phi is either one or it is k plus one  if it's one then of course f must be of degree k plus one otherwise f will be linear  so let us see the two cases we will address first the case when  degree of phi of x equals k plus one we just recall that this is how fk and g of x are  composed now from this composition we see that the degree of f must be one because this is of  degree k plus one phi is of degree k plus one so f must be linear but then clearly we have fk of  x equals phi t of x where t of x is f lambda of x it's linear now we can take the inverse function  of the polynomial function t of x so that t of t minus one gives the identity function  so this means that fk of t minus x is just phi of x so we can express phi of x in the form  fk of something which is linear now if we go to the other part of the  or other polynomial g of x g of x is composed like phi of g of tappa of x  but phi can be written like fk of t minus one and now you see that this is fk of something  and in principle we don't care what is here inside the only important feature is that it is a polynomial  with rational coefficients and of degree at least one so this means that we can prove that g of x  equals fk of h of x this is the exceptional case when we may have infinitely many solutions  and we will see that there are no other possibilities for infinitely many solutions  since the other possibility is when the degree of phi of x equals one whenever the degree of phi  of x equals one we can write phi of x like this and in this case we have to study all the possible  standard pairs now we have five kinds of standard pairs in the case of standard pairs of type two  this can be excluded because there we have a polynomial of degree two but f of x and g of x  cannot be of degree two in this case because f of x must be of degree k plus one if i go back  here now phi is linear lambda is linear so in order to get the polynomial of degree k plus one  f must be of degree k plus one it cannot be quadratic polynomial so on the other hand  the other polynomial let me just come back for a second here to the second type standard pairs  so one of them must be x square and the other one must be of this shape  but here it's clear that this polynomial has multiple roots whenever the degree is larger than  two or larger than larger equal to four sorry so in that case again we see that it has multiple  roots but f k of x cannot have multiple roots so  this way we can exclude polynomials of pairs of polynomials of type two standard pairs of type two  standard pairs of certain force kind they are the polynomials which are defined using  with some polynomials we can exclude them by comparing coefficients in this equation  in the case of standard pairs of type five in principle we have two cases f k of x must be  either of this shape which contradicts with the in the composibility of f if we look at this  then this shows that there is a polynomial eight times lambda x square minus one to the cube  but this means that this polynomial is composed to phi x cube plus phi one x cube plus phi zero  so this is a non-trivial composition of polynomials f k cannot be of this shape  on the other hand we may have f k of x equals this shape and in this case we just compare again  coefficients and after comparing three four coefficients we get a contradiction  in the case of the first type again the root structure of f k of x and the derivative of  f k of x prove that such standard pairs are impossible well of course here we have some  lengthier computation which I have no time to to show for these situations but this is the flavor  of the of the proof we also proved an effective result  in the case of the equation f k of x equals a y to the n plus b  we could prove that the solutions are bounded by an effectively computable constant  and well this is a generalization of the result of Hoidl-Lichram and Tange  who considered the case a equals one and b equals zero now we will switch our attention  to a different topic. Comornic polynomials are polynomials introduced by Riemann-Kommornig  for some results concerning digital expansions and if we define a sub n of x equals x to the power  two times three to the power n minus x to the power three to the n minus one and we multiply  such polynomials from k equals zero to n then we get the n's Komornig polynomial.  Komornig polynomials the first few of them look like this the first one is easy to see  but the second one and the third one are these polynomials so we can see that these are polynomials  with coefficients plus or minus one and there are the degree of this of such polynomials is always  c to the power n plus one minus one. All coefficients are plus or minus one below the highest degree  term below the main term there are no other coefficients than plus or minus one including  zero there is no zero coefficient so all the terms are present it consists of c to the power n plus  one monomials and all these monomials have coefficients either plus one or minus one.  It is also proved that all the complex roots of such polynomials are simple.  The first important result concerning Komornig polynomials is that Komornig polynomials are  incomposable over c and this helps us to use this for to apply the B-Lutih C-R-M later  but for the moment let us see a few more properties of a k of x. In principle there is a polynomial  identity a l plus k of x equals a of a sub l of x to the power three to the power k and if we denote  the Fibonacci numbers but by phi zero, pi one and so on and we define  f sub k to be phi of three to the power k plus one plus phi of three to the k minus one  minus one then these are clearly integer numbers these integer numbers will play an important  role in the GCD of two values of this polynomial a k of x. So if we take the polynomial a k of x  in any integer space and we compute it for the index l and for the index l plus k then the GCD of  these two integers will divide the integer a sub k defined just here using the Fibonacci numbers.  So this way for polynomials with index close to each other we can compute the GCD or we can  prove how much is the GCD if they are consecutive in this is k and k plus one  then the GCD will always be one and if between the indexes between the indices the difference is two  then the GCD is either one or five.  So after these auxiliary properties let us show about show the results concerning  the of anti-equations containing Pomornik polynomials the problem originates from Atilla Petu  who asked in a conference about the finiteness of number of solutions of e sub n of x equals e  sub n of y. With Kolushpink and Schustek we did prove that the equation p sub n of x equals p sub m  of y has always finitely many integer solutions and even more general we consider the equation with  arbitrary right hand side and we did prove that if g of x is not of the form  p sub n of h of x or g of x is not of the form gamma of delta x to the l where gamma and delta are  linear polynomials directional coefficients and l is just the degree of this polynomial g  then we have only finitely many solutions. However I have to mention that the first exception that's  necessary the second exception we couldn't prove that in this case there are infinitely many solutions  so it may it probably may be excluded but we were just unable to prove that.  Well that was what it was written here and of course since we use BDT-CRM the result is ineffective.  We also did prove some results concerning this polynomial which are effective in the case  of the equation p of n of x equals y to the m. We did get effective solutions both in the case  m equals to or m larger would be equal to 3 and we even could prove an effective result  but for the exponent m from here whenever y is different of plus or minus one and for the  concrete cases of y square in the right hand side we did solve completely this equation  there are some trivial solutions and except of the trivial solutions it exists only one  non-trivial solution in the case when n equals 2 and we did solve the equation also for n equals 1,  2, 3, 4 and 5. Further we did prove we did solve completely the equation p of n  of x equals y cube for n equals 0, 1, 2 and 3. As my time is going I will not tell about  these results in details however I will switch to my third topic power values of power sums.  Well if you take s k of x 1 to the k plus 2 to the k plus x to the k this is s k of x and the  problem s k of x equals y to the n this is a well-known and much-investigated classical  problem it goes back to Luca Watson and the first international known base through was you to  check for could prove that if k and n is not one of these pairs then this equation 16 has  only 590 many solutions. He also conjectured that if k and n are not k and is not of this  pair not not one of these pairs then the only non-trivial solution is given by 2, 2, 24 and 70.  Many results concerning this equation have been established during the last few decades.  Yuri Tiedemann and Forkhoof proved an effective version of Schaeffer and they have another extra  feature that also n was an unknown so in this equation also the exponent was an unknown for them  and the effective finiteness was working also in this more general case. Then Pinter proved that  for non-trivial solutions we have this bound and the conjecture was verified by Jacobson Pinter  and Walsh for some bounded values of k and n equals 2. Benadio and Pinter extended the result to  one smaller or equal to k smaller or equal to 11 and arbitrary and larger or equal to 2. This was  again an important breakthrough and Pinter verified the conjecture for and larger than 4 and one  smaller or equal to k smaller than 107. This is of course not a complete list of results.  However, how do we initiate the investigation of this equation for fixed values of x? For values  of x which are smaller than 25 and are congruent of 0 and 3 mod 4, he computed all the solutions and  together with Koidu Mierzak and Inc. we extended this result for the cases when k is congruent to  one or two mod 4. In principle this shows that we proved the Schaeffer conjecture for x smaller  than 25 and larger or equal to 3. Now finally I would like to mention  few other kinds of equations where S k of x is included and the Bielow-Tihe CRM is used,  but these results are not mine. I couldn't contribute to this topic. Bielow-Grinsow-Kinscher  Hofer-Pieter and Tihe consider the equation S k of x equals SM of y and for the case when  Geofi is the product of consecutive integers Bielow-Binser Hofer-Pieter and Tihe proved again,  then Jürj Kovac, Peter and Inter for binomial coefficients and for this kind of sums Jürj  Kovac, Peter and Inter. So also for such kind of separable equations Bielow-Tihe CRM was used  and it was a powerful tool. Thank you very much for your attention. This was what I wanted to tell.  Yeah so thank you very much Attila for your very nice and rich talk. So it's open for discussion  and I already can see a very precise question by Gary Walsh who is our next speaker. So the question  you can read here is it possible to read it? Just a second I'm trying to find out. So he is asking  where do the upper pounds come on page 28? On page 28. Very precise question.  Page 28 just let me find oh no I'm sorry no no it's 20 oh  yeah 28. I see well these are coming from super and hyper elliptic equations from Baker's theory  from a result of Jahn de Kewer, Czekalman, Jürj and myself. We have a  an explicit a completely explicit result for this case and we use that CRM for this.
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