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Gökhan Soydan

Publications and source records attributed to Gökhan Soydan.

At least 19 recordsLinked to original sources
arXiv2026

Perfect powers in sequences of polygonal numbers

Let $P_s(n)$ denote the $n$-th $s$-gonal number. Consider the Diophantine equation $P_{s}(n) = t^{m}$ for integers $n, s, t$ and $m > 2$. All solutions to this equation are known for $m>2$ and $s\in\{3,5,6,8,10,20\}$. Here we extend these results to the cases $s = 2k+4$ (where $k = 4,6$ or $5 \leq k \leq 97$ is a prime number) and $s = k+4$ (where $k = 9,15$ or $3 \leq k \leq 97$ is a prime number). The proofs of our results use the modular and hypergeometric methods, linear forms in logarithms and extensive calculations. We were unable to completely solve the above Diophantine equations, but we expect (based on GRH and the weak effective $abc$ conjecture) that there will be no additional solutions beyond those explicitly shown in Theorems~1, 2 and 3.

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arXiv2023

On the Ternary Purely Exponential Diophantine Equation $(ak)^x+(bk)^y=((a+b)k)^z$ with Prime Powers $a$ and $b$

Let $k$ be a positive integer, and let $a,b$ be coprime positive integers with $\min\{a,b\}>1$. In this paper, using a combination of some elementary number theory techniques with classical results on the Nagell-Ljunggren equation, the Catalan equation and some new properties of the Lucas sequence (\seqnum{A000204} in OEIS), we prove that if $k>1$ and $a,b$ are both prime powers with $\min\{a,b\}>2$, then the equation $(ak)^x+(bk)^y=((a+b)k)^z$ has only one positive integer solution $(x,y,z)=(1,1,1)$. The above result partially proves that Conjecture 1 presented in (Acta Arith. 2018, 184 (1): 37-49) is true.

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arXiv2022

A modular approach to the generalized Ramanujan-Nagell equation

Let $k$ be a positive integer. In this paper, using the modular approach, we prove that if $k\equiv 0 \pmod{4}$, $30< k<724$ and $2k-1$ is an odd prime power, then under the GRH, the equation $x^2+(2k-1)^y=k^z$ has only one positive integer solution $(x,y,z)=(k-1,1,2)$. The above results solve some difficult cases of Terai's conecture concerning this equation.

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arXiv2021

On the power values of the sum of three squares in arithmetic progression

In this paper, using a deep result on the existence of primitive divisors of Lehmer numbers due to Y. Bilu, G. Hanrot and P. M. Voutier, we first give an explicit formula for all positive integer solutions of the Diophantine equation $(x-d)^2+x^2+(x+d)^2=y^n$ (*) when $n$ is an odd prime and $d=p^r$, $p>3$ a prime. So this improves the results on the papers of A. Koutsianas and V. Patel \cite{KP} and A. Koutsianas \cite{Kou}. Secondly, under the assumption of our first result, we prove that (*) has at most one solution $(x,y)$. Next, for a general $d$, we prove the following two results: (i) if every odd prime divisor $q$ of $d$ satisfies $q\not\equiv \pm 1 \pmod{2n},$ then (*) has only the solution $(x,y,d,n)=(21,11,2,3)$. (ii) if $n>228000$ and $d>8\sqrt{2}$, then all solutions $(x,y)$ of (*) satisfy $y^n<2^{3/2}d^3$.

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arXiv2021

On a class of generalized Fermat equations of signature $(2,2n,3)$

We consider the Diophantine equation $7x^{2} + y^{2n} = 4z^{3}$. We determine all solutions to this equation for $n = 2, 3, 4$ and $5$. We formulate a Kraus type criterion for showing that the Diophantine equation $7x^{2} + y^{2p} = 4z^{3}$ has no non-trivial proper integer solutions for specific primes $p > 7$. We computationally verify the criterion for all primes $7 < p < 10^9$, $p \neq 13$. We use the symplectic method and quadratic reciprocity to show that the Diophantine equation $7x^{2} + y^{2p} = 4z^{3}$ has no non-trivial proper solutions for a positive proportion of primes $p$. In the paper \cite{ChDS} we consider the Diophantine equation $x^{2} +7y^{2n} = 4z^{3}$, determining all families of solutions for $n=2$ and $3$, as well as giving a (mostly) conjectural description of the solutions for $n=4$ and primes $n \geq 5$.

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arXiv2021

The Shuffle Variant of a Diophantine equation of Miyazaki and Togbé

In 2012, T. Miyazaki and A. Togbé gave all of the solutions of the Diophantine equations $(2am-1)^x+(2m)^y=(2am+1)^z$ and $b^x+2^y=(b+2)^z$ in positive integers $x,y,z,$ $a>1$ and $b\ge 5$ odd. In this paper, we propose a similar problem (which we call the shuffle variant of a Diophantine equation of Miyazaki and Togbé). Here we first prove that the Diophantine equation $(2am+1)^x+(2m)^y=(2am-1)^z$ has only the solutions $(a, m, x, y, z)=(2, 1, 2, 1, 3)$ and $(2,1,1,2,2)$ in positive integers $a>1,m,x,y,z$. Then using this result, we show that the Diophantine equation $b^x+2^y=(b-2)^z$ has only the solutions $(b,x, y, z)=(5, 2, 1, 3)$ and $(5,1,2,2)$ in positive integers $x,y,z$ and $b$ odd.

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arXiv2021

On elliptic curves induced by rational Diophantine quadruples

In this paper, we consider elliptic curves induced by rational Diophantine quadruples, i.e. sets of four nonzero rationals such that the product of any two of them plus 1 is a perfect square. We show that for each of the groups $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/k\mathbb{Z}$ for $k = 2, 4, 6, 8$, there are infinitely many rational Diophantine quadruples with the property that the induced elliptic curve has this torsion group. We also construct curves with moderately large rank in each of these four cases.

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arXiv2020

Rational Points in Geometric Progression on the Unit Circle

A sequence of rational points on an algebraic planar curve is said to form an $r$-geometric progression sequence if either the abscissae or the ordinates of these points form a geometric progression sequence with ratio $r$. In this work, we prove the existence of infinitely many rational numbers $r$ such that for each $r$ there exist infinitely many $r$-geometric progression sequences on the unit circle $x^2 + y^2 = 1$ of length at least $3$.

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arXiv2020

On the exponential Diophantine equation $(n-1)^{x}+(n+2)^{y}=n^{z}$

Suppose that $n$ is a positive integer. In this paper, we show that the exponential Diophantine equation $$(n-1)^{x}+(n+2)^{y}=n^{z},\ n\geq 2,\ xyz\neq 0$$ has only the positive integer solutions $(n,x,y,z)=(3,2,1,2), (3,1,2,3)$. The main tools on the proofs are Baker's theory and Bilu-Hanrot-Voutier's result on primitive divisors of Lucas numbers.

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arXiv2020

A note on the ternary purely exponential Diophantine equation $A^x+B^y=C^z$ with $A+B=C^2$

Let $\ell, m, r$ be fixed positive integers such that $2\nmid \ell$, $3\nmid \ell m$, $\ell>r$ and $3\mid r$. In this paper, using the BHV theorem on the existence of primitive divisors of Lehmer numbers, we prove that if $\min\{r\ell m^2-1,(\ell-r)\ell m^2+1\}>30$, then the equation $(r\ell m^2-1)^x+((\ell -r)\ell m^2+1)^y=(\ell m)^z$ has only the positive integer solution $(x,y,z)=(1,1,2)$.

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arXiv2020

On the Diophantine equation $(5pn^{2}-1)^{x}+(p(p-5)n^{2}+1)^{y}=(pn)^{z}$

Let $p$ be a prime number with $p>3$, $p\equiv 3\pmod{4}$ and let $n$ be a positive integer. In this paper, we prove that the Diophantine equation $(5pn^{2}-1)^{x}+(p(p-5)n^{2}+1)^{y}=(pn)^{z}$ has only the positive integer solution $(x,y,z)=(1,1,2)$ where $pn \equiv \pm1 \pmod 5$. As an another result, we show that the Diophantine equation $(35n^{2}-1)^{x}+(14n^{2}+1)^{y}=(7n)^{z}$ has only the positive integer solution $(x,y,z)=(1,1,2)$ where $n\equiv \pm 3% \pmod{5}$ or $5\mid n$. On the proofs, we use the properties of Jacobi symbol and Baker's method.

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arXiv2020

Resolution of the equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$

Consider the diophantine equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$ in positive integers $x_1\le x_2$, and $y_1\le y_2$. Each side of the equation is a product of two terms of a given binary recurrence, respectively. In this paper, we prove that the only solution to the title equation is $(x_1,x_2,y_1,y_2)=(1,2,1,1)$. The main novelty of our result is that we allow products of two terms on both sides.

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arXiv2020

On a class of Lebesgue-Ljunggren-Nagell type equations

Given odd, coprime integers $a$, $b$ ($a>0$), we consider the Diophantine equation $ax^2+b^{2l}=4y^n$, $x, y\in\Bbb Z$, $l \in \Bbb N$, $n$ odd prime, $\gcd(x,y)=1$. We completely solve the above Diophantine equation for $a\in\{7,11,19,43,67,163\}$, and $b$ a power of an odd prime, under the conditions $2^{n-1}b^l\not\equiv \pm 1(\mod a)$ and $\gcd(n,b)=1$. For other square-free integers $a>3$ and $b$ a power of an odd prime, we prove that the above Diophantine equation has no solutions for all integers $x$, $y$ with ($\gcd(x,y)=1$), $l\in\mathbb{N}$ and all odd primes $n>3$, satisfying $2^{n-1}b^l\not\equiv \pm 1(\mod a)$, $\gcd(n,b)=1$, and $\gcd(n,h(-a))=1$, where $h(-a)$ denotes the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-a})$.

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arXiv2019

The Diophantine Equation $(x+1)^k+(x+2)^k+\cdots+(\ell x)^k=y^n$ Revisited

Let $k,\ell\geq2$ be fixed integers and $C$ be an effectively computable constant depending only on $k$ and $\ell$. In this paper, we prove that all solutions of the equation $(x+1)^{k}+(x+2)^{k}+...+(\ell x)^{k}=y^{n}$ in integers $x,y,n$ with $x,y\geq1, n\geq2, k\neq3$ and $\ell\equiv 1 \pmod 2$ satisfy $\max\{x,y,n\}<C$. The case when $\ell$ is even has already been completed by Soydan (Publ. Math. Debrecen 91 (2017), pp. 369-382).

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arXiv2019

Rational sequences on different models of elliptic curves

Given a set $S$ of elements in a number field $k$, we discuss the existence of planar algebraic curves over $k$ which possess rational points whose $x$-coordinates are exactly the elements of $S$. If the size $|S|$ of $S$ is either $4,5$, or $6$, we exhibit infinite families of (twisted) Edwards curves and (general) Huff curves for which the elements of $S$ are realized as the $x$-coordinates of rational points on these curves. This generalizes earlier work on progressions of certain types on some algebraic curves.

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arXiv2018

Elliptic Curves Containing Sequences of Consecutive Cubes

Let $E$ be an elliptic curve over $\mathbb{Q}$ described by $y^2= x^3+ Kx+ L$ where $K, L \in \mathbb{Q}$. A set of rational points $(x_i,y_i) \in E(\mathbb{Q})$ for $i=1, 2, \cdots, k$, is said to be a sequence of consecutive cubes on $E$ if the $x-$coordinates of the points $x_i$'s for $i=1, 2, \cdots$ form consecutive cubes. In this note, we show the existence of an infinite family of elliptic curves containing a length-$5$-term sequence of consecutive cubes. Morever, these five rational points in $E (\mathbb{Q})$ are linearly independent and the rank $r$ of $E(\mathbb{Q})$ is at least $5$.

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