On the Diophantine equation $\binom{n}{k}=\binom{m}{l}+d$
Homero R. Gallegos-Ruiz, Nikolaos Katsipis, Szabolcs Tengely and, Maciej Ulas
TL;DR
This paper completely solves a class of Diophantine equations involving binomial coefficients and a small integer shift by analyzing integral points on elliptic and hyperelliptic curves, providing new computational and theoretical insights.
Contribution
It introduces a method to solve the binomial coefficient Diophantine equation for specific parameters by finding all integral points on related algebraic curves, expanding understanding of these equations.
Findings
Complete solutions for the specified binomial equations with -3 ≤ d ≤ 3.
Identification of all integral points on certain elliptic and hyperelliptic curves.
Additional computational and theoretical observations related to the equations.
Abstract
By finding all integral points on certain elliptic and hyperelliptic curves we completely solve the Diophantine equation for and Moreover, we present some other observations of computational and theoretical nature concerning the title equation.
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Taxonomy
TopicsAlgebraic Geometry and Number Theory · Analytic Number Theory Research · Cryptography and Residue Arithmetic
On the Diophantine equation
H. R. Gallegos-Ruiz
Homero R. Gallegos-Ruiz
Unidad Académica de Matemáticas
Universidad Autónoma de Zacatecas
Calzada Solidaridad y Paseo de la Bufa
Zacatecas, Zacatecas, CP 98000
Mexico
,
N. Katsipis
Nikolaos Katsipis
Department of Mathematics & Applied Mathematics
University of Crete
GR-70013, Heraklion, Crete
Greece
,
Sz. Tengely
Szabolcs Tengely
Institute of Mathematics
University of Debrecen
P.O.Box 12
4010 Debrecen
Hungary
and
M. Ulas
Maciej Ulas
Jagiellonian University
Faculty of Mathematics and Computer Science
Institute of Mathematics
Łojasiewicza 6
30-348 Kraków
Poland
and
Institute of Mathematics of the Polish Academy of Sciences
Świȩtego Tomasza 30
31-014 Kraków, Poland
Abstract.
By finding all integral points on certain elliptic and hyperelliptic curves we completely solve the Diophantine equation for and Moreover, we present some other observations of computational and theoretical nature concerning the title equation.
Key words and phrases:
binomial coefficient, Diophantine equation, elliptic curve, genus two curve, integer points
2000 Mathematics Subject Classification:
Primary 11G30, Secondary 11J8
1. Introduction
There are many nice results related to the equation
[TABLE]
in unknowns , , , This is usually considered with the restrictions and The only known solutions (with the above mentioned restrictions) are the following
[TABLE]
where is the th Fibonacci number. The infinite family of solutions involving Fibonacci numbers was found by Lind [17] and Singmaster [21].
Equation (1) has been completely solved for pairs
[TABLE]
In cases of these pairs one can easily reduce the equation to the determination of solutions of a number of Thue equations or elliptic Diophantine equations. In 1966, Avanesov [1] found all integral solutions of equation (1) with De Weger [10] and independently Pintér [19] provided all the solutions of the equation with The case reduces to the equation which was solved by Mordell [18]. The remaining pairs were handled by Stroeker and de Weger [27], using linear forms in elliptic logarithms. The case with was completely solved by Bugeaud, Mignotte, Siksek, Stoll and Tengely [8], the integral solutions are as follows
[TABLE]
In a recent paper Blokhuis, Brouwer and de Weger [4] determined all non-trivial solutions with or General finiteness results are also known. In 1988, Kiss [15] proved that if and is a given odd prime, then the equation has only finitely many positive integral solutions. Using Baker’s method, Brindza [6] showed that equation (1) with and has only finitely many positive integral solutions.
In case of the more general equation
[TABLE]
Blokhuis, Brouwer and de Weger [4] determined all non-trivial solutions with and and They provided a complete list of solutions for the above cases and if
\begin{array}[]{ll}\small\begin{tabular}[]{|l|l|l|l|}\hline\crn&k&m&l\\ \hline\cr\hline\cr11&2&8&3\\ \hline\cr60&2&23&3\\ \hline\cr160403633&2&425779&3\\ \hline\cr6&3&7&2\\ \hline\cr7&3&9&2\\ \hline\cr16&3&34&2\\ \hline\cr27&3&77&2\\ \hline\cr29&3&86&2\\ \hline\cr34&3&21&4\\ \hline\cr\end{tabular}&\small\begin{tabular}[]{|l|l|l|l|}\hline\crn&k&m&l\\ \hline\cr\hline\cr19630&3&1587767&2\\ \hline\cr12&4&32&2\\ \hline\cr93&4&2417&2\\ \hline\cr10&5&23&2\\ \hline\cr22&5&230&2\\ \hline\cr62&5&3598&2\\ \hline\cr135&5&26333&2\\ \hline\cr139&5&28358&2\\ \hline\cr28&11&6554&2\\ \hline\cr\end{tabular}\end{array}
Table 1. Known solutions of the Diophantine equation .
If is not fixed they also obtained some interesting infinite families, an example is given by
[TABLE]
In 2019, Katsipis [14] completely resolved the case with and he also determined the integral solutions if and
The aim of this paper is to extend results mentioned above and offer some general observations and computational results.
2. Main results
We start our discussion with some numerical observations. More precisely, we observed that for certain pairs and an integer , the congruence
[TABLE]
with suitable chosen prime number , has no solutions. This immediately implies unsolvability in integers of the related Diophantine equation.
Theorem 1**.**
If and is a quadratic non-residue modulo , where the -adic valuation of is odd, then congruence (3) has no solutions. In particular, equation (2) has no solutions in integers.
Remark*.*
Based on the previous theorem we may provide some explicit results, for example if where then equation (2) has no solutions in integers with
By using elementary number theory we compute all integral solutions of equation (2) for some values of and with and We note that the case is in some sense trivial. Indeed, in this case the solvability of equation (5) is equivalent to the existence of integers such that and . Equivalently, we need to determine integers with and satisfying the conditions
[TABLE]
Thus, if is odd, one can take , where , i.e., the number of solutions of our equation is at least , where . If is even one possible choice is .
Theorem 2**.**
All integral solutions of equation (2) with and are as follows
[TABLE]
In the next result we deal with the cases that can be reduced to elliptic curves.
Theorem 3**.**
All integral solutions of equation (2) with and are as follows.
[TABLE]
\begin{array}[]{ll}\small\par\begin{tabular}[]{|c|l|}\hline\crd&(k,l)=(2,4)\\ \hline\cr\hline\cr3&\left[\right]\\ \hline\cr2&\left[(4,3)\right]\\ \hline\cr1&\left[(5,4),(7,9),(12,32),(93,2417)\right]\\ \hline\cr0&\left[(4,2),(6,6),(10,21)\right]\\ \hline\cr-1&\left[\right]\\ \hline\cr-2&\left[\right]\\ \hline\cr-3&\left[\right]\\ \hline\cr\end{tabular}&\small\begin{tabular}[]{|c|l|}\hline\crd&(k,l)=(2,6)\\ \hline\cr\hline\cr3&\left[(7,5),(11,31),(50,5638)\right]\\ \hline\cr2&\left[(6,3)\right]\\ \hline\cr1&\left[\right]\\ \hline\cr0&\left[(6,2),(8,8),(10,21),(14,78)\right]\\ \hline\cr-1&\left[(7,4)\right]\\ \hline\cr-2&\left[\right]\\ \hline\cr-3&\left[\right]\\ \hline\cr\end{tabular}\end{array}**
\begin{array}[]{ll}\small\begin{tabular}[]{|c|l|}\hline\crd&(k,l)=(2,8)\\ \hline\cr\hline\cr3&\left[\right]\\ \hline\cr2&\left[(8,3)\right]\\ \hline\cr1&\left[(5,9),(32,12)\right]\\ \hline\cr0&\left[(8,2),(10,10),(14,78),(17,221)\right]\\ \hline\cr-1&\left[\right]\\ \hline\cr-2&\left[\right]\\ \hline\cr-3&\left[(9,4)\right]\\ \hline\cr\end{tabular}&\small\begin{tabular}[]{|c|l|}\hline\crd&(k,l)=(3,4)\\ \hline\cr\hline\cr3&\left[(4,4)\right]\\ \hline\cr2&\left[\right]\\ \hline\cr1&\left[\right]\\ \hline\cr0&\left[(4,3),(7,7)\right]\\ \hline\cr-1&\left[(5,4),(21,34)\right]\\ \hline\cr-2&\left[\right]\\ \hline\cr-3&\left[\right]\\ \hline\cr\end{tabular}\end{array}**
\begin{array}[]{lll}\begin{tabular}[]{|c|l|}\hline\crd&(k,l)=(3,6)\\ \hline\cr\hline\cr3&\left[(6,4),(7,5)\right]\\ \hline\cr2&\left[\right]\\ \hline\cr1&\left[\right]\\ \hline\cr0&\left[(6,3),(9,9)\right]\\ \hline\cr-1&\left[\right]\\ \hline\cr-2&\left[\right]\\ \hline\cr-3&\left[(7,4)\right]\\ \hline\cr\end{tabular}&\begin{tabular}[]{|c|l|}\hline\crd&(k,l)=(4,6)\\ \hline\cr\hline\cr3&\left[\right]\\ \hline\cr2&\left[\right]\\ \hline\cr1&\left[\right]\\ \hline\cr0&\left[(6,4),(10,10)\right]\\ \hline\cr-1&\left[\right]\\ \hline\cr-2&\left[(7,5)\right]\\ \hline\cr-3&\left[\right]\\ \hline\cr\end{tabular}&\begin{tabular}[]{|c|l|}\hline\crd&(k,l)=(4,8)\\ \hline\cr\hline\cr3&\left[\right]\\ \hline\cr2&\left[\right]\\ \hline\cr1&\left[\right]\\ \hline\cr0&\left[(8,4),(12,12)\right]\\ \hline\cr-1&\left[\right]\\ \hline\cr-2&\left[\right]\\ \hline\cr-3&\left[\right]\\ \hline\cr\end{tabular}\end{array}**
Among the solutions given by Blokhuis, Brouwer and de Weger [4] there are some with e.g.:
[TABLE]
in these cases the problem can be reduced to genus 2 curves.
Theorem 4**.**
All integral solutions of equation (2) with are as follows.
[TABLE]
Let be odd. In the following theorem we consider the Diophantine equation
[TABLE]
in polynomials satisfying the condition . Note that if is a solution of (4), then due to the identity , is also a solution. In the sequel we count such pairs of solutions as one. We are motivated by findings presented in [4].
Theorem 5**.**
Let be a variable.
- (1)
For equation (4) has exactly three solutions. 2. (2)
For equation (4) has exactly one solution. 3. (3)
For equation (4) has no solutions.
3. Proofs of the theorems
Proof of Theorem 1.
In order to get the result it is enough to note that the equation can be rewritten as
[TABLE]
where . If , then . Under our assumption on we see that 3 is quadratic non-residue modulo and congruence (3), and hence equation (2), has no integer solutions. ∎
Motivated by the result above, we performed numerical search for pairs and prime numbers such that the congruence (3) has no solutions modulo . Here are results of our computations.
[TABLE]
Table 2. Pairs such that there exist such that for some the congruence (3) has no solutions.
Proof of Theorem 2.
Here we obtain that
[TABLE]
and the polynomial is reducible. It follows that
[TABLE]
Hence divides It remains to solve the one variable polynomial equation
[TABLE]
for ∎
Remark*.*
Let us note that if , then in the considered range, i.e, we have found at most one integer solution. It is an interesting problem to look for values of such that the equation
[TABLE]
has more than one solution in positive integers satisfying . In order to construct values of such that equation (5) has “many” solutions we used the following strategy. First, we computed the set
[TABLE]
and then looked for duplications in . We considered . As one could expect, in the case the number of duplicates is big. In fact, we found 488 values of which appeared at least three times in . The smallest value correspond to with the solutions . We found only three values of such that equation (5) has four solutions. The values of and the corresponding solutions are as follows:
[TABLE]
We strongly believe that the following is true.
Conjecture**.**
For each there is such that the equation has at least positive integer solutions.
For we found 1190 values of which appeared at least two times in . The smallest value corresponds to with the solutions . We found only one value of such that equation (5) has three solutions. More precisely, for equation (5) has three solutions .
For we found 4 values of which appeared at least 2 times in . The values of and the corresponding solutions are as follows:
[TABLE]
For we also found 4 values of which appeared at least 2 times in . The values of and the corresponding solutions are as follows:
[TABLE]
For we found only one value of such that equation has two solutions. For we have solutions .
For there are no duplicates in the set .
Proof of Theorem 3.
All the equations related to this part can be reduced to elliptic curves given is some model.
[TABLE]
Table 3. Elliptic models of certain Diophantine equations of the form
There exists a number of software implementations for finding integral points on elliptic curves [5, 22]. These procedures are based on a method developed by Stroeker and Tzanakis [28] and independently by Gebel, Pethő and Zimmer [13]. One may follow the transformations provided in [27] to handle these cases. Here we used the Magma procedures IntegralPoints() and IntegralQuarticPoints(). In some cases there exist no solution and we used IsLocallySolvable() and TwoCoverDescent() [7]. In cases related to we follow the above mentioned elliptic logarithm method, the cases with were solved earlier as given in the introduction, so it remains to deal with the values
The case yields an elliptic curve with Mordell-Weil rank 3 while the remaining three values of yield elliptic curves with Mordell-Weil rank 2; we only provide details for the case .
For this case we set , and we have the equation
[TABLE]
where and and the Weierstrass model which is birationally equivalent to over is
[TABLE]
A notation remark: We will use “exponents” C and E on a point to declare whether the point is viewed as one on or , respectively. Also, we will use or for the -coordinates or the -coordinates, respectively.
As already mentioned, has rank 3; its free part is generated by the points
[TABLE]
and the torsion subgroup is trivial.
The birational transformation between the models and is
[TABLE]
with
[TABLE]
[TABLE]
where
[TABLE]
and
[TABLE]
With the aid of Maple we find out that there is exactly one conjugacy class of Puiseux series solving . This unique class contains exactly three series and only the following one has real coefficients:
[TABLE]
Here is the cubic root of . For every real solution of with it is true that (according to Lemma 8.3.1 in [29]).
Then the point that plays a crucial role in the resolution (see [29, Definition 8.3.3]) is
[TABLE]
Referring to the discussion of Section 1 of [14], we consider the linear form
[TABLE]
Since has only one real root, namely , we have , therefore coincides with the elliptic logarithm of for (see Chapter 3 of [29], especially, Theorem 3.5.2). On the other hand, has irrational coordinates. As Magma does not possess a routine for calculating elliptic logarithms of non-rational points, we wrote our own routine in Maple for computing -values of points with algebraic coordinates.
Thus we compute
[TABLE]
[TABLE]
Note that the four points are -linearly independent because their regulator is non-zero (see [20, Theorem 8.1]). Therefore our linear form falls under the scope of the second “bullet” in [29, page 99] and we have , , , , for , , , , and , so that, in the relation (9.6) of [29] we can take
[TABLE]
We compute the canonical heights of using Magma111For the definition of the canonical height we follow J.H. Silverman; as a consequence the values displayed here for the canonical heights are the halves of those computed by Magma and the least eigenvalue of the height-pairing matrix below, is half that computed by Magma; cf. “Warning” at bottom of p. 106 in [29].
and for the canonical height of we confine ourselves to the upper bound by applying [29, Proposition 2.6.4]. Thus we have
[TABLE]
[TABLE]
The corresponding height-pairing matrix for the particular Mordell-Weil basis is
[TABLE]
with minimum eigenvalue
[TABLE]
Next we apply [29, Proposition 2.6.3] in order to compute a positive constant with the property that for every point , where denotes Weil height;222In the notation of [29, Proposition 2.6.3], as a curve we take the minimal model of which is itself. it turns out that
[TABLE]
Finally, we have to specify the constants defined in [29, Theorem 9.1.2]. This can be carried out almost automatically with a Maple program. In this way we compute
[TABLE]
According to [29, Theorem 9.1.3], applied to “case of Theorem 8.7.2”, if , where and are explicit positive constants, then either , where is an explicit constant, or
[TABLE]
where all constants involved in it are explicit. More specifically, (in a similar way as in Appendix B in [14] for the case of ),
[TABLE]
So, in view of (14) and (10), (11), (12), (13), we conclude that, if , then either or
[TABLE]
But for all , we check that the left-hand side is strictly larger than the right-hand side which implies that , therefore
[TABLE]
An easy straightforward computation shows that is the only one integer point with (equivalently, the integer solution of (6) with ).
In order to find explicitly all points with it is necessary to reduce the huge upper bound (15) to an upper bound of manageable size. This is accomplished with LLL-algorithm [16], in a similar way as in Appendix D in [14], and we obtain the reduced bound . Therefore, we have to check which points
[TABLE]
have the property that maps via the transformation (LABEL:UVd2) to a point with integer coordinates. We remark here that every point with integer and is obtained in this way, but the converse is not necessarily true; i.e. if and the above maps to with integer coordinates, it is not necessarily true that . After a computational search we find the only one point which corresponds to the zero point .
So no integral solution (with and ) of equation (2) with exists.
For the other three cases we provide some details in the tables below:
[TABLE]
Table 4. and
[TABLE]
Table 5. Upper bounds of .
[TABLE]
Table 6. All points with .
∎
Proof of Theorem 4.
We provide details only in case of here the rank of the Jacobian is 6 (like in case of ). Equation (2) with defines the hyperelliptic curve
[TABLE]
Based on Stoll’s papers [23], [24], [25] one can determine generators for the Mordell-Weil group by using Magma [5]. We obtain that is free of rank with Mordell-Weil basis given by (in Mumford representation)
[TABLE]
and the torsion subgroup is trivial. We apply Baker’s method [2] to get a large upper bound for here we use the improvements given in [8] and [12]. It follows that
[TABLE]
We have from Corollary 3.2 of [12] that every integral point on the curve can be expressed in the form
[TABLE]
with Proposition 6.2 in [12] gives an estimate for the precision we need to compute the appropriate matrices, this bound is as follows
[TABLE]
where in our case and We choose to compute the period matrix and the hyperelliptic logarithms with 1500 digits of precision. The hyperelliptic logarithms of the divisors are given by
[TABLE]
We need now to choose an integer that is larger than the constant given by Proposition 6.2 in [12]. Setting we get a new bound for We repeat the reduction process with that yields a better bound, namely Two more steps with and provide the bounds and It remains to compute all possible expressions of the form
[TABLE]
with We performed a parallel computation to enumerate linear combinations coming from integral points on a machine having 12 cores. The computation took 3 hours and 23 minutes. We obtained the following non-trivial solutions
[TABLE]
If then the rank of the Jacobian is 6 and the Baker bound is and we have that In three steps it is reduced to In this case the non-trivial solutions are as follows
[TABLE]
If then the rank of the Jacobian is 3, we followed the arguments given in [8] and [11] to obtain a large bound for the size of possible integral solutions. We present them in the table below.
[TABLE]
Table 7. Upper bounds for .
In all three cases the rank of the Jacobians are equal to 3 and the torsion subgroup is trivial hence all points can be written as
[TABLE]
where Using the previously applied hyperelliptic logarithm method the initial large upper bounds for can be significantly reduced. If then after one reduction step we get the bound 64 and other two steps make it 7. The only pair of integral points we obtain is given by Therefore we have
[TABLE]
If then first we obtain a reduced bound 51 and finally it follows that The complete list of integral points is given by Thus we obtain
[TABLE]
Finally, in case of the first reduction yields a bound 58 and the third one provides 6. The complete set of integral solutions is so we do not get non-trivial solution of (2).
If then the rank of the Jacobian is 1, therefore classical Chabauty’s method [9] can be applied, it is now implemented in Magma [5]. We obtain that the equation has no non-trivial solution. ∎
Remark*.*
Let
[TABLE]
and write . The curve is isomorphic to the curve defined by the equation . We computed upper bounds for the numbers using the Magma procedure RankBound. We obtained the following data
[TABLE]
Table 8. Upper bounds for the rank of Jacobian of the curve for .
We checked that for the upper bounds computed by RankBound are actually equal to the ranks.
Let us note that and . We checked that in both cases the rank is equal to 7. This follows from the existence of seven independent divisors in . They are as follows:
[TABLE]
We also looked for high rank Jacobians for further values of of the form For we obtained the equality with the following independent divisors
[TABLE]
The torsion part of is trivial. We conjecture that the only solutions in positive integers of the equation are
[TABLE]
The large points are explained by the fact that on the curve we have the following solutions
[TABLE]
Hence we obtain the following divisors on
[TABLE]
Remark*.*
In case of the equation
[TABLE]
one obtains genus 3 curves. Stoll [26] proved that the rank of the Jacobian is 9 if For other values of in the range many of the genus 3 hyperelliptic curves have high ranks as well. Balakrishnan et. al. [3] developed an algorithm to deal with genus 3 hyperelliptic curves defined over whose Jacobians have Mordell-Weil rank 1. If then the equation is isomorphic to the curve
[TABLE]
and using Magma (with SetClassGroupBounds("GRH") to speed up computation) we get that the rank of the Jacobian is 1. Therefore we may try to use the Sage implementation described in [3] to compute the set of rational points on this curve. The affine points are hence we have the solution
[TABLE]
Proof of Theorem 5.
In each case we will be working in the same way. More precisely, for given we write and . The polynomial needs to be zero. Thus the coefficient near in need to be zero for . In consequence, we are interested in solving the system of polynomial equations
[TABLE]
in variables . We have and thus for some non-zero . We note that after the substitution of the computed values of into the system , the related system of equations
[TABLE]
where comes from after the substitution of the computed values of , is triangular with respect to the variables . More precisely, we have for . Moreover, the coefficient near is a power of times a rational number. Solving for and substituting into we are left with the system of equations
[TABLE]
in three variables . The polynomial is the numerator of the rational function after substitution of the computed values . It seems that for each fixed odd , the system can be solved using Gröbner bases techniques. More precisely, we compute - the Gröbner basis of the ideal generated by the polynomials . For we have more equations than variables we expect that the system for all sufficiently large has no rational (and even complex) solutions. This can be confirmed with our approach for . However, we were unable to prove such a statement in full generality.
We prove the first part of our theorem. However, we present details of the reasoning only for . The case is proved in exactly the same way. We are interested in rational solutions of the system
[TABLE]
We have for some . We put the values of into the system and solve corresponding system of equations
[TABLE]
with respect to . We get
[TABLE]
In consequence, after the substitution of the values of into the system we obtain the system
[TABLE]
where . It is an easy task to solve the system . Indeed, we compute Gröbner basis , of the ideal generated by . The basis contains four polynomials. Two of them are the following
[TABLE]
and we easily obtain the following solutions
[TABLE]
Note that the first two solutions were presented in [4]. Unfortunately, the polynomials from the third solution take only non-integer values.
For we proceed in the same way and omit details. However, let us note that the Gröbner basis contains 7 polynomials. Two of them are the following
[TABLE]
and we obtain two solutions with integer coefficients and the solution (corresponding to the triple )
[TABLE]
By replacing by we obtain polynomial with integer coefficients, which is exactly the third solution from the paper [4].
For the Gröbner basis contains 11 elements. In particular, the following three polynomials are in :
[TABLE]
We found that the only solution (corresponding to ) is the following
[TABLE]
The last part of our theorem follows from certain Gröbner basis computations. For we found that the contains polynomial of the form for some , i.e., need to be zero which leads to contradiction. ∎
Remark*.*
Using the same approach as in the proof of the above theorem one can prove that the Diophantine equation has no polynomial solutions satisfying for .
We also looked for solutions of the more general Diophantine equation
[TABLE]
where is of degree 2. By using the same approach as in the proof of Theorem 5 one can prove that for there are no solutions of (16) satisfying and .
However, if we allow to be of degree 3 we found the following solutions. For we have the solution
[TABLE]
For we have the solution
[TABLE]
Note that in both cases by replacing by we get polynomials with integer coefficients.
Playing around with the Diophantine equation we also found the polynomial solution
[TABLE]
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