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 ...
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
$$ x^y y^z z^x = x^z y^x z^y. \tag{$\...
81
votes
Estimating the size of solutions of a diophantine equation
This exact problem is the subject of the paper "An Unusual Cubic Representation Problem" by Andrew Bremner (ASU) and myself. It was published in Volume 43 (2014) of Annales Mathematicae et ...
41
votes
Is $x^2+x+1$ ever a perfect power?
Your equation can be rewritten as
$$\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&...
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 $X$, ...
34
votes
Accepted
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 ...
34
votes
Accepted
Can $9xy$ divide $1+x^2+x^3+y^2$?
The equation is solvable in integers. Take, for example,
$$
x = -19578556686240310295378317903565, \\
y = -101658411567714319887, \\
z = 418962851513108789978912616277180591709694.
$$
Verification can ...
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 ...
31
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 ...
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 ...
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 ...
29
votes
Accepted
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
$$
ap^x + bq^y = c+ dp^z q^...
28
votes
What is the smallest unsolved Diophantine equation?
I will concentrate on the first question: ''One does not know the integral solutions of $P(x)=0$.'' To avoid discussion what exactly is meant by ''know the solutions'' if there are infinitely many of ...
28
votes
Can $x^4+y^4+1$ be a perfect power?
To answer question 2:
$$346^4+36788^4+1=1831575032204939793=3^3\cdot19^3\cdot179^2\cdot17569^2.$$
27
votes
Infinitely many integer solutions to $X^4+Y^4-18Z^4= -16$
I do not know if the result is trivial or known, but your result is fun
and follows from the identity
$$(x-1)^4+(x+1)^4+16=2(x^2+3)^2$$
so that you find infinitely many solutions by solving the Pell ...
27
votes
Accepted
Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$?
Yes, it is true that $\{ x^{4} + y^{2} + z^{2} : x, y, z \in \mathbb{Z}[i] \} = \{ a + 2bi : a, b \in \mathbb{Z} \}$. Indeed, one can even take $x$ to be either $0$ or $1$ in all cases. Because $y^{2}+...
27
votes
Accepted
Is "almost-solvability" of Diophantine equations decidable?
A Diophantine equation is almost-satisfiable iff it is satisfiable over the ring $\widehat{\mathbb Z}$, the profinite completion of $\mathbb Z$ (also called by some the Prüfer ring), by a standard ...
Community wiki
26
votes
Accepted
When does $axy+byz+czx$ represent all integers?
Here is a proof of the conjecture. I will refer several times to the book Cassels: Rational quadratic forms (Academic Press, 1978).
1. Let $p$ be a prime such that $p\nmid a$. Using the invertible ...
25
votes
Accepted
Does $2^x-3p^y=5$ (with $p$ an odd prime) have only finitely many positive integer solutions?
The solutions of your equation can be injected into the solutions of the $S$-unit equation over $\mathbb{Q}$, where $S=\{\infty,2,3,5,p\}$. As the latter is known to have finitely many solutions by ...
25
votes
Why does representing functors help solving Diophantine equations?
E.g. let $f(x,y, z)=0$ be a smooth projective plane curve with $f$ a rational polynomial of degree $\ge 4$. Then Mordell conjectured, and Faltings proved, that this has only finitely many rational ...
24
votes
Diophantine equation $3^n-1=2x^2$
A standard (although possibly not the most efficient) way to solve equations of this sort is to find all integer points on each of the elliptic curves
$$ E_1: X^3-1=2Y^2,\quad E_2:3X^3-1=2Y^2,\quad ...
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 ...
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 (...
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 ...
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 ...
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 $t,k$ ...
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 $4(xy+1)(xy+x+2)=4n^2=(2xy+x+3-z)^2$ ...
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 \...
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) ...
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