128 votes
Accepted

Estimating the size of solutions of a diophantine equation

This problem turned out to be much more interesting than I originally thought. Let me give my solution, which seems to be slightly different from (but essentially the same as) the solution in the ...
Michael Stoll's user avatar
87 votes
Accepted

About the validity of a new conjecture about a diophantine equation

The conjecture is true, in fact the equation has no solution in distinct positive real numbers. To see this, let us write the equation in the more symmetric form $$ x^y y^z z^x = x^z y^x z^y. \tag{$\...
GH from MO's user avatar
  • 96.9k
81 votes

Estimating the size of solutions of a diophantine equation

This exact problem is the subject of the paper "An Unusual Cubic Representation Problem" by Andrew Bremner (ASU) and myself. It was published in Volume 43 (2014) of Annales Mathematicae et ...
Allan MacLeod's user avatar
41 votes

Is $x^2+x+1$ ever a perfect power?

Your equation can be rewritten as $$\frac{x^{3}-1}{x-1} = y^{N}.$$ As Gerhard Paseman commented above, the Diophantine equation $$ \frac{x^{n} − 1}{x-1} = y^{q} \quad x > 1, \quad y>1, \quad n&...
José Hdz. Stgo.'s user avatar
35 votes
Accepted

Is equation $xy(x+y)=7z^2+1$ solvable in integers?

There is no solution. It is clear that at least one of $x$ and $y$ is positive and that neither is divisible by 7. We can assume that $a := x > 0$. The equation implies that there are integers $X$, ...
Michael Stoll's user avatar
34 votes
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Universality of $y^4-x^3$ mod $p$

The curve $C:y^4-x^3z=az^4$ is nonsingular over $\mathbb F_p$ for $p\ge5$ and $a\ne0$. It has genus $3$. So Weil's theorem says that $$ \bigl| \#C(\mathbb F_p) - p - 1 \bigr| \le 6\sqrt{p}. $$ There ...
Joe Silverman's user avatar
33 votes

Permutations $\pi\in S_n$ with $\sum_{k=1}^n\frac1{k+\pi(k)}=1$

Claim: $a_n>0$ for all $n\geq 6\quad (*)$. Proof: We use induction to prove $(*)$. We have $a_6,a_7,a_8,a_9,a_{10},a_{11}>0$. Assume $(*)$ holds for all the integers $\in [6,n-1]$. We want ...
Zhao Shen's user avatar
  • 331
30 votes
Accepted

The "stubborn" solutions to sums of three cubes

This lim sup indeed goes to $\infty$. We can prove this using exactly the strategy Lucia suggested. We will count the number of $x,y,z$ in a box with $x^3+y^3+z^3$ not a cubic residue modulo $p$ for a ...
Will Sawin's user avatar
  • 133k
30 votes

Can you solve the listed smallest open Diophantine equations?

The equation $$ x^3 + y^3 + z^3 + xyz = 5 $$ is solvable in integers. For example, take $$ x=-3028982, \quad y=-3786648, \quad z=3480565. $$ Verification is straightforward, but I would like to add ...
Bogdan Grechuk's user avatar
30 votes
Accepted

Can $y^2-4$ be a divisor of $x^3-x^2-2 x+1$?

No. The roots of $x^3 - x^2 - 2x + 1$ are $-(\zeta + \zeta^{-1})$ where $\zeta$ is a 7th root of unity; this soon implies [see below] that any prime factor is either $7$ or $\pm 1 \bmod 7$, and thus ...
Noam D. Elkies's user avatar
29 votes
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Diophantine equation $3^a+1=3^b+5^c$

I can't resist this: The young Chris Skinner showed that if $a$, $b$, $c$, $d$ are fixed positive integers, and $p$ and $q$ are positive coprime integers then the equation $$ ap^x + bq^y = c+ dp^z q^...
Lucia's user avatar
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28 votes

What is the smallest unsolved Diophantine equation?

I will concentrate on the first question: ''One does not know the integral solutions of $P(x)=0$.'' To avoid discussion what exactly is meant by ''know the solutions'' if there are infinitely many of ...
Bogdan's user avatar
  • 743
27 votes

Infinitely many integer solutions to $X^4+Y^4-18Z^4= -16$

I do not know if the result is trivial or known, but your result is fun and follows from the identity $$(x-1)^4+(x+1)^4+16=2(x^2+3)^2$$ so that you find infinitely many solutions by solving the Pell ...
Henri Cohen's user avatar
  • 11.3k
27 votes
Accepted

Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$?

Yes, it is true that $\{ x^{4} + y^{2} + z^{2} : x, y, z \in \mathbb{Z}[i] \} = \{ a + 2bi : a, b \in \mathbb{Z} \}$. Indeed, one can even take $x$ to be either $0$ or $1$ in all cases. Because $y^{2}+...
Jeremy Rouse's user avatar
27 votes

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

To answer question 2: $$346^4+36788^4+1=1831575032204939793=3^3\cdot19^3\cdot179^2\cdot17569^2.$$
Thomas Browning's user avatar
27 votes
Accepted

Is "almost-solvability" of Diophantine equations decidable?

A Diophantine equation is almost-satisfiable iff it is satisfiable over the ring $\widehat{\mathbb Z}$, the profinite completion of $\mathbb Z$ (also called by some the Prüfer ring), by a standard ...
26 votes
Accepted

When does $axy+byz+czx$ represent all integers?

Here is a proof of the conjecture. I will refer several times to the book Cassels: Rational quadratic forms (Academic Press, 1978). 1. Let $p$ be a prime such that $p\nmid a$. Using the invertible ...
GH from MO's user avatar
  • 96.9k
25 votes
Accepted

Does $2^x-3p^y=5$ (with $p$ an odd prime) have only finitely many positive integer solutions?

The solutions of your equation can be injected into the solutions of the $S$-unit equation over $\mathbb{Q}$, where $S=\{\infty,2,3,5,p\}$. As the latter is known to have finitely many solutions by ...
GH from MO's user avatar
  • 96.9k
25 votes

Why does representing functors help solving Diophantine equations?

E.g. let $f(x,y, z)=0$ be a smooth projective plane curve with $f$ a rational polynomial of degree $\ge 4$. Then Mordell conjectured, and Faltings proved, that this has only finitely many rational ...
Donu Arapura's user avatar
  • 34.1k
24 votes

Diophantine equation $3^n-1=2x^2$

A standard (although possibly not the most efficient) way to solve equations of this sort is to find all integer points on each of the elliptic curves $$ E_1: X^3-1=2Y^2,\quad E_2:3X^3-1=2Y^2,\quad ...
Joe Silverman's user avatar
24 votes
Accepted

Diophantine equation $3^n-1=2x^2$

This problem happens to have appeared on the Polish Mathematical Olympiad camp in 2015. Here is the official solution of the problem: (I use $m$ in place of $x$ because this is how the problem was ...
Wojowu's user avatar
  • 26.9k
24 votes

Why is this "the first elliptic curve in nature"?

I actually only wrote the part that says that this curve is a model for $X_1(11)$, not the first part, which I think was written by John Cremona. It is standard to order elliptic curves by conductor (...
Nicolas Mascot's user avatar
23 votes
Accepted

Rational inscribed realization of the regular dodecahedron

An example Yes, here is a list of rational coordinates lying on the unit sphere, the convex hull of which is combinatorially equivalent to a regular dodecahedron. This polyhedron is invariant under ...
Adam P. Goucher's user avatar
23 votes

How many cubes are the sum of three positive cubes?

There are bivariate coprime polynomial parametrizations: https://sites.google.com/site/tpiezas/010: $$(a^4-2ab^3)^3 + (a^3 b+b^4)^3 + (2a^3 b-b^4)^3 = (a^4+a b^3)^3.$$ Added If you drop the positivity ...
joro's user avatar
  • 24.2k
23 votes
Accepted

Are there infinitely many positive integer solutions to $(3+3k+l)^2=m\,(k\,l-k^3-1)$?

It does have infinitely many positive solutions. Here is just one such series. Consider the following recurrence sequence: $$u_0=1,\ u_1=2,\ u_{n+1} = 23 u_n - u_{n-1} - 4\qquad (n\geq 1).$$ Let $t,k$ ...
Max Alekseyev's user avatar
22 votes

Does the equation $(xy+1)(xy+x+2)=n^2$ have a positive integer solution?

It looks that Vieta jumping helps. For fixed positive integer $y$ choose a minimal positive integer $x$ for which $(xy+1)(xy+x+2)$ is a perfect square. Denote $4(xy+1)(xy+x+2)=4n^2=(2xy+x+3-z)^2$ ...
Fedor Petrov's user avatar
22 votes
Accepted

Simplest diophantine equation with open solvability

Determining which integers $n$ are a sum of three cubes is a very famous open problem: $$a^3 + b^3 + c^3 = n, \quad a,b,c \in \mathbb{Z}.$$ Conjecturally, $n$ is a sum of three cubes iff $n \not \...
Daniel Loughran's user avatar
21 votes

What is the smallest unsolved Diophantine equation?

It might be enlightening to divide equations into further types - first three decidable types, then four undecidable types. (1) Equations which have a small or otherwise easy-to-find solution (2) ...
Will Sawin's user avatar
  • 133k
21 votes
Accepted

Is the diophantine equation $3x^2+1=py^2$ always solvable for each prime $p\equiv 13\pmod{24}$?

Updated on 2019/08/21: I prove Conjectures 1-3 below. I will use the usual terminology and notations for binary quadratic forms. In particular, I will use the first two pages of Pall: Discriminantal ...
GH from MO's user avatar
  • 96.9k

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