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8 votes
0 answers
245 views

Hilbert 10th problem for genus 2 equations

Hilbert 10th problem, while undecidable in general, remains open for 2-variable equations: we do not know if there is an algorithm that, for polynomial $P(x,y)$ with integer coefficients, decides ...
Bogdan Grechuk's user avatar
11 votes
1 answer
540 views

Prove that $1$ is the sum of three tetrahedral numbers infinitely many different ways

It's well known that $1$ is the sum of three cubes infinitely many different ways but is it true for perhaps the tetrahedral numbers as well? Let $T_n = (1/6)n(n+1)(n+2)$. Then the following are the ...
Benjamin L. Warren's user avatar
2 votes
1 answer
360 views

Positive divisors of $P(x,n)=1+x+x^2+ \cdots + x^n$ that are congruent to $1$ modulo $x$

This is a follow-up question to Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$ Let \begin{equation} P(x,n)= 1+x+x^2+ \cdots + x^n, \end{equation} \begin{...
ASP's user avatar
  • 319
5 votes
0 answers
215 views

Integer points of rational function in 2 variables

Is there an algorithm that, given polynomials $P(x)$ and $Q(y)$ with integer coefficients, decides whether there exists integers $x$ and $y$ such that $\frac{P(x)}{Q(y)}$ is an integer? This is a ...
Bogdan Grechuk's user avatar
0 votes
1 answer
87 views

Diophantine equations that involve Lehmer means with all digits equal to $1$ in their $x-$adic expansions

In this post I present my variations of the problem involving Nagell-Ljunggren equation, that is explained in pages 10 and 11 of Highlights in the Research Work of T. N. Shorey by R. Tijdeman, from ...
user142929's user avatar
6 votes
2 answers
743 views

Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$

Let $P(z) = z^4 +z^3 +z^2 +z+1$ where $z$ is a positive integer. While working with the diophantine equation $(xz+1)(yz+1)=P(z)$, I was able to construct a seemingly infinite and complete solution set ...
ASP's user avatar
  • 319
0 votes
0 answers
138 views

A diophantine equation involving partial sums of exponentials similar than the equation in Fermat's Last Theorem

I'm curious about the following diophantine equation from my invention: I don't know if this is in the literature, I wrote it using creativity in an attempt to write a variant of the equation in ...
user142929's user avatar
6 votes
1 answer
357 views

Can $2^n\pm n$ with $n>2$ be a triangular number?

Recall that triangular numbers are those $$T(n)=\frac{n(n+1)}2\ \ (n=0,1,2,\ldots\}.$$ Clearly, $$2^1-1=1=T(1),\ \ 2^1+1=3=T(2),\ \ 2^2+2=6=T(3).$$ Question. Is there an integer $n>2$ with $2^n-n$ ...
Zhi-Wei Sun's user avatar
  • 15.6k
9 votes
1 answer
702 views

Software for detecting Brauer-Manin obstructions?

In the context of another MO question, the following question arose: Does there exist any software for detecting Brauer–Manin obstructions to the existence of integer solutions to a single polynomial ...
Timothy Chow's user avatar
  • 82.7k
6 votes
2 answers
896 views

How can I find all integer solutions of $3^n - x^2 = 11$

I know that $n$ can't be even because of the following argument: Let $n = 2p$. Then we can use the difference of two squares and it becomes like this : $(3^p + x)(3^p - x) = 11; 3^p + x = 11 , 3^p - x ...
İsmim Yok's user avatar
21 votes
1 answer
1k views

Is "almost-solvability" of Diophantine equations decidable?

Say that a Diophantine equation is almost-satisfiable iff for each $n\in\mathbb{N}$ it has a solution mod $n$. Trivially genuine satisfiability over $\mathbb{N}$ implies almost-satisfiability, but the ...
Noah Schweber's user avatar
7 votes
1 answer
880 views

A family of Diophantine equations with no integer solutions but solutions modulo every integer

Selmer's curve is the equation $3x^3 +4y^3 +5z^3=0$. This equation is famous for having non-trivial solutions in every completion of $\mathbb{Q}$ but only having the trivial solution in the rationals. ...
JoshuaZ's user avatar
  • 6,969
72 votes
3 answers
8k views

Can you solve the listed smallest open Diophantine equations?

In 2018, Zidane asked What is the smallest unsolved Diophantine equation? The suggested way to measure size is substitute 2 instead of all variables, absolute values instead of all coefficients, and ...
Bogdan Grechuk's user avatar
3 votes
1 answer
262 views

Determine if a 2-variable Diophantine equation has a finite or infinite number of solutions

Do there exist an algorithm, which, given a polynomial $P(x,y)$ with integer coefficients, determines whether Diophantine equation $P(x,y)=0$ has finite or infinite number of integer solutions? Famous ...
Bogdan's user avatar
  • 781
1 vote
1 answer
203 views

On the Diophantine equation $m^2 - p^k = 2^r t$, where $r \geq 2$ and $\gcd(2,t)=1$

This question is an offshoot of this closely related MO question. Here, we consider the Diophantine equation $$m^2 - p^k = 2^r t,$$ where $r \geq 2$ and $\gcd(2,t)=1$. Furthermore, we place the ...
Jose Arnaldo Bebita's user avatar
6 votes
3 answers
606 views

Non-trivial solution to $\sum^{n}_{i=1}\sum^{n}_{j=1,j\ne i}(x_{i})^{(x_j)}=(\sum^{n}_{i=1}x_i)^{(\sum^{n}_{i=1}x_i)}$

This problem was first asked at Mathematics Stack Exchange, where it wasn't drawn much attention. For ease of reading, $$S=\sum_{i=1}^nx_i, S_p=\sum_{i=1,i\ne p}^nx_i, S^{[q]}=\sum_{i=1}^nx_i^q, S_p^{[...
Alex-Github-Programmer's user avatar
9 votes
0 answers
563 views

Iterating Diophantine equations over Q to quickly get a large interval with just integer solutions

Hilbert's Tenth Problem was whether there is an algorithm which will answer whether any Diophantine equation has solutions (where we want integer solutions). Hilbert's Tenth has a negative solution by ...
JoshuaZ's user avatar
  • 6,969
10 votes
4 answers
1k views

The smallest solution to $2^{2k}-1=\text{powerful}$

Integer is powerful if all the exponents in its factorization are at least $2$. Every powerful integer can be written in the form $a^2 b^3$. For odd $k$, define $F(k)=2^{2k}-1=(2^k-1)(2^k+1)$. This ...
joro's user avatar
  • 25.4k
22 votes
2 answers
1k views

Why is 1331 the only cube of the form $x^2 + x − 1$?

The Wikipedia (https://en.wikipedia.org/wiki/1000_(number)#1300_to_1399) mentions that 1331 is the only cube of the form $x^2 + x − 1$, for $x = 36$. What is the proof?
Tanya Khovanova's user avatar
0 votes
0 answers
112 views

The number of solutions of $2^xpx+k=y^2$

Let's consider the family of diophantine equations $$2^xpx+k=y^2$$ being $p\gt2$ a prime and $k$ a positive integer. An example is given by the equation $$2^x\cdot3x+97=y^2$$ that presents, at least, ...
Augusto Santi's user avatar
1 vote
2 answers
435 views

On the equation $x^3 + y^3 =cz^3$

What are the characteristics of the values of $c$ for which the equation $x^3 + y^3 = cz^3$ has pairwise coprime non-zero integral solutions where $x \neq \pm y$ ? For instance, it is known that $c$ ...
Otto's user avatar
  • 11
7 votes
2 answers
911 views

Triangular numbers of the form $x^4+y^4$

Recall that triangular numbers are those $T(n)=n(n+1)/2$ with $n\in\mathbb N=\{0,1,2,\ldots\}$. Fermat ever proved that the equation $x^4+y^4=z^2$ has no positive integer solution. So I think it's ...
Zhi-Wei Sun's user avatar
  • 15.6k
8 votes
1 answer
639 views

On Markoff-type diophantine equation

Do there exist integers $x,y,z$ such that $$ x^2+y^2-z^2 = xyz -2 \quad ? $$ Why this is interesting? First, this equation arose in an answer to the previous Mathoverflow question What is the smallest ...
Bogdan's user avatar
  • 781
8 votes
1 answer
806 views

Can $x^4+y^4+1$ be a perfect power?

Recall that a perfect power has the form $x^m$ with $m,x\in\{2,3,\ldots\}$. Motivated by Fermat's result that the equation $x^4+y^4=z^2$ has no positive integer solution, here I ask the following ...
Zhi-Wei Sun's user avatar
  • 15.6k
3 votes
0 answers
135 views

Will an integer combination of some number of copies of the set of powers of 2 and the set of powers of 3 always have natural density 0?

Consider a Diophantine equation of the form $$(c_1 2^{x_1} + \dots + c_n 2^{x_n}) + (c_{n+1} 3^{x_{n+1}} + \dots + c_m 3^{x_m}) = y$$ where $x_1, \dots, x_m, y$ are our variables (here $x_1, \dots, ...
Christian Schulz's user avatar
1 vote
0 answers
161 views

On the equations $(x^n+1)(y^n+1)=z^2+1$ and $(x^n-1)(y^n-1)=z^2+1$

Note that $$(1^2+1)(2^2+1)=10=3^2+1\ \ \mbox{and}\ \ (1^4+2^4)(5^4+6^4)=8^4+13^4.$$ Today I tried to find positive integers $x,y,z$ satisfying $(x^4+1)(y^4+1)=z^4+1$ but failed. In view of this ...
Zhi-Wei Sun's user avatar
  • 15.6k
3 votes
1 answer
233 views

Pythagorean triples and quadratic residues modulo primes

QUESTION. Are my following conjectures true? How to prove them? Conjecture 1. For each prime $p>100$, there are $a,b,c\in\{1,\ldots,p-1\}$ such that $$\left(\frac ap\right)=\left(\frac bp\right)=\...
Zhi-Wei Sun's user avatar
  • 15.6k
6 votes
1 answer
332 views

Is $xz+1 $ a proper divisor of $a_3z^3+a_2z^2+a_1z+1$ finitely often?

Given a polynomial $P=a_3z^3+a_2z^2+a_1z+1, z >0$ with non-negative integer coefficients $a_1, a_2, a_3\ne 0$, it appears if $P$ is not factorizable then there are finitely many positive integers $...
ASP's user avatar
  • 319
10 votes
0 answers
217 views

Are the nonnegative rationals diophantine with only two quantifiers?

Definition: A subset $D\subseteq \mathbb{Q}$ is diophantine if it is the projection of the zero set of a polynomial, i.e. there exists a polynomial $f\in\mathbb{Q}[X,Y_1,\dots,Y_n]$ for some $n$ such ...
Arno Fehm's user avatar
  • 2,051
3 votes
0 answers
198 views

Positive integers $m$ such that $2m^2-1=x^4+y^4$ for some $x,y\in\{0,1,\ldots\}$ with $|x-y|>1$

I note that $$2(n^2+n+1)^2 -1= n^4+(n+1)^4.$$ This leads me to pose the following question. Question 1. Are there infinitely many positive integers $m$ such that $2m^2-1=x^4+y^4$ for some $x,y\in\...
Zhi-Wei Sun's user avatar
  • 15.6k
4 votes
0 answers
238 views

Diophantine consequences of the Buzzard–Diamond–Jarvis conjecture

Serre's modularity conjecture famously implies Fermat Last Theorem. More generally, Serre's conjecture implies that certain generalized Fermat equations have no non-trivial solutions (see Section 4.1 ...
Tristan Phillips's user avatar
4 votes
1 answer
571 views

Relation between stacky curves and "M-curves"

A tame stacky curve over a field $k$ is a geometrically connected proper smooth DM stack of dimension 1 which has a dense open substack which is a scheme, and whose automorphism group of each ...
k.j.'s user avatar
  • 1,364
7 votes
0 answers
274 views

Is decidability reducible to unique decidability (perhaps in multilinear polynomial situations)?

Given a Diophantine equation it is not decidable if it has integer solution. I. Is there a Diophantine set $\mathcal D_{unique}$ satisfying the properties every member in $\mathcal D_{unique}$ is a ...
Turbo's user avatar
  • 13.9k
-2 votes
1 answer
494 views

Why wolfram alpha gives integers solutions for some equations of the form $ x^3 +(k\times10^n)^3 + z^3=0 $?

I have tried to get representations of some integers as sum of three cubic of the form $x^3+(k*10^n)^3+z^3$ with $k$ is integer and $n$ is a postive integer, I took this example : $(48807585839879)^3-(...
zeraoulia rafik's user avatar
3 votes
0 answers
308 views

Is there an integer-valued quadratic polynomial $P(x,y)$ such that $\{P(x,y)+2^k:\ x,y\in\mathbb Z\ \text{and}\ k\in\mathbb N\}=\mathbb Z^+$?

I seek for very sparse representations of positive integers. Let $$\mathbb N=\{0,1,2,\ldots\}\ \ \ \text{and}\ \ \ \mathbb Z^+=\{1,2,3,\ldots\}.$$ Recall that a polynomial $P(x,y)$ is integer-valued ...
Zhi-Wei Sun's user avatar
  • 15.6k
4 votes
0 answers
176 views

Natural numbers in the form $\lfloor\frac{a^3+b^3}2+\frac{c^3+d^3}6\rfloor$

Let $\mathbb N=\{0,1,2,\ldots\}$. Several years ago I proved that $$\{aw^3+bx^3+cy^3+dz^3:\ w,x,y,z\in\mathbb N\}\not=\mathbb N$$ for any positive integers $a,b,c,d$ (cf. http://maths.nju.edu.cn/~...
Zhi-Wei Sun's user avatar
  • 15.6k
0 votes
1 answer
228 views

On the equation $y^2 = x^3 - z^3$ [closed]

What is the parametric form of the rational solutions of the equation $y^2 = x^3 - z^3 ?$
PRIMES is in P.'s user avatar
4 votes
1 answer
273 views

Is $\prod_{k=1}^nk^{\sigma(k)}$ a square or a cube for some $\sigma\in S_n$?

Note that for any permutation $\sigma\in S_5$ the product $\prod_{k=1}^5k^{\sigma(k)}$ is neither a square nor a cube. Question. Let $n>5$ be an integer. Is the product $\prod_{k=1}^nk^{\sigma(k)}$ ...
Zhi-Wei Sun's user avatar
  • 15.6k
0 votes
0 answers
266 views

Is there a permutation $\tau\in S_n$ with $\tau(1)^{\tau(2)}+\cdots+\tau(n-1)^{\tau(n)}+\tau(n)^{\tau(1)}$ a square?

Let $n>1$ be an integer, and let $S_n$ be the symmetric group of all the permutatins of $\{1,\ldots,n\}$. I'm curious whether there is a permutation $\tau\in S_n$ such that $$\tau(1)^{\tau(2)}+\...
Zhi-Wei Sun's user avatar
  • 15.6k
3 votes
0 answers
116 views

Variation in decidability of diophantine equations with field extension

Let $O_k$ be the ring of integers in a subfield $k$ of $\overline{\mathbb{Q}}$. Let's call an equation $p(x_1, \dots, x_n) = 0$ where $p$ is a polynomial in $n$-variables $x_1, \dots, x_n$ with ...
Fanta's user avatar
  • 31
2 votes
0 answers
266 views

On the equations $x^yy^z=z^x$ and $w^x+x^y+y^z=z^w$

Recently, I considered the equation $$x^yy^z=z^x\qquad(x,y,z\in\{2,3,\ldots\}).\tag{1}$$ The equation $(1)$ has infinitely many solutions with $x=z$ including $$(x,y,z)=(n^n,n^{n-1},n^n),\ (n^{2n^2},n^...
Zhi-Wei Sun's user avatar
  • 15.6k
8 votes
4 answers
2k views

Status of $x^3+y^3+z^3=6xyz$

In Erik Dofs, Solutions of $x^3 + y^3 + z^3 = nxyz$, Acta Arithmetica 73 (1995) pp. 201–213, doi:10.4064/aa-73-3-201-213, EuDML the author has studied the Diophantine equation \begin{equation} x^3+y^...
Haran's user avatar
  • 371
5 votes
1 answer
345 views

Quadratic Diophantine equations with all values prime

Given a quadratic Diophantine equation over the integers in two variables, can we say much about when it has only finitely many solutions with the additional assumption that both variables are prime? ...
JoshuaZ's user avatar
  • 6,969
1 vote
0 answers
102 views

Finding number fields over which Diophantine equations are solvable

Given a Diophantine equation $f(x_1, \dots, x_n) \in \mathbb{Z}[x_1, \dots, x_n]$ and a family of number fields $K$ (say, the number fields of a specified degree and signature), are there techniques ...
bean's user avatar
  • 479
6 votes
3 answers
876 views

Is there a simple proof that $Ax^3+By^3=C$ has only finitely many integer solutions

One can use Thue's 1909 result to show that the Diophantine equation $Ax^3 + By^3 = C$ ($A,B$ not perfect cubes, $C\neq 0$) has finitely many integer solutions $(x,y)$. But does there exist a simple ...
math110's user avatar
  • 4,280
26 votes
1 answer
2k views

The "stubborn" solutions to sums of three cubes

It is conjectured (see [1]) that for any integer $k\not\equiv \pm 4\pmod 9$ there are infinitely many integer solutions to $$ a^3+b^3+c^3=k. $$ Numerical investigations of this conjecture show that ...
Alexander Kalmynin's user avatar
4 votes
1 answer
271 views

$n$ variables Diophantine

Let $n \ge 2$ be a positive integer. Do there exist $n$ non-zero distinct integers such that the sum of their square is a perfect square and their product is a nth power? For $n=2$ the answer is no, ...
jack's user avatar
  • 3,153
-2 votes
2 answers
148 views

Mordell like equation [closed]

This looks like a mordell like equation X²=Y³-25056 How to solve it? The exact equation is (36x)²=(6y)³-25056 Is there any website has records of the equation x²=y³+k For k>25000
sonu's user avatar
  • 1
1 vote
0 answers
216 views

How to solve special Diophantine equation systems (which one can solve by hand) with the computer?

I have a quadratic Diophantine equation system which is possibly not homogeneous and has some mixed and some linear terms. But I know that there are only finitely many solutions over the integers. One ...
Bernhard Boehmler's user avatar
4 votes
3 answers
286 views

Finding Pythagorean quadruples on a given plane?

In 2D one cannot construct Pythagorean triples $x^2+y^2=m^2$ ($x,y,m\in\mathbb{Z}$) that lie on every line through the origin (e.g., a Pythagorean triple with $x=y$ would require $\sqrt{2}$ to be ...
Jim McCann's user avatar

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