118
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 ...

86
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{$\...

73
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 ...

40
votes

Accepted

### Rational points on the "quintic circle" $x^5 + y^5 = 7$

There is an action of $\mu_5$, the group of fifth roots of unity, on your curve,
given by $\zeta \cdot (x,y) = (\zeta x, \zeta^{-1} y)$. The quotient by this
group action is the hyperelliptic curve
$$...

39
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 ...

38
votes

Accepted

### Does the equation $241+2^{2s+1}=m^2$ have a solution?

To answer your first question: there is indeed no $s$ such that
$241+2^{2s+1}$ is a perfect square. -- Proof: $2^{2s+1}$ is always
congruent to either $2$, $8$ or $32$ modulo $63$, which makes
$241+2^{...

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 ...

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 ...

33
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 $X$, ...

31
votes

Accepted

### When is $(q^k-1)/(q-1)$ a perfect square?

The equation
$$ \frac{x^k-1}{x-1}=y^m$$
is known as the Nagell-Ljunggren equation. It is conjectured that for $x\geq 2$, $y\geq 2$, $k\geq 3$, $m\geq 2$, the only solutions are
$$ \frac{3^5-1}{3-1}=...

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 ...

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

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 ...

27
votes

### Which integers can be expressed as a sum of three cubes in infinitely many ways?

As stated by Dietrich Burde, it is known that there is no solution for $n\equiv \pm 4 \pmod{9}$, and conjectured that there are infinitely many solutions otherwise.
A cryptic aspect is that it is not ...

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

### 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 ...

26
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 ...

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

### On a result attributed to W. Ljunggren and T. Nagell

I went to my office today and scanned the Ljunggren's paper that OP asked for. I provide some bibliographical information first:
Wilhelm Ljunggren, Noen setninger om ubestemte likninger av formen $...

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 ...

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

### 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

### 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 ...

23
votes

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

To answer question 2:
$$346^4+36788^4+1=1831575032204939793=3^3\cdot19^3\cdot179^2\cdot17569^2.$$

23
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 ...

22
votes

### Fermat's last theorem over larger fields

This is not quite an answer, but not quite a comment either.
We can at least show that $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$ is infinite (where $X$ is the quintic Fermat curve). There are in ...

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

### Simplest diophantine equation with open solvability

The perfect cuboid problem: We do not know if there is a common integer solution to
$$a^2+b^2+c^2=d^2$$
$$a^2+b^2=e^2$$
$$a^2+c^2=f^2$$
$$b^2+c^2=g^2$$
with $a,b,c \ge 1$.
The last condition can ...

22
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 ...

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