107

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 paper by Bremner and
MacLeod (see Allan MacLeod's answer).
Theorem. Let $a,b,c$ be positive integers. Then
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ can ...

75

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
$$ x^y y^z z^x = x^z y^x z^y. $$
We get the same equation after interchanging $x$ and $y$, or $y$ and $z$, i.e., after permuting the variables arbitrarily. Hence we can assume without loss of ...

64

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 Informaticae, pages 29-41.
It is proven that strictly positive solutions never exist for $n$ odd. They sometimes do not exist for $n$ even, and, even if they do, they can ...

40

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
$$C \colon Y^2 = X^5 + \frac{49}{4},$$
the map being given by $(X, Y) = (-xy, x^5 - \frac{7}{2})$. So it is enough
to find all the rational points on $C$. $C$ is ...

38

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
$241+2^{2s+1}$ congruent to either $21$, $54$ or $60$ modulo $63$.
However none of these values is a quadratic residue modulo $63$,
and thus $241+2^{2s+1}$ cannot be a ...

37

Your equation can be rewritten as
$$\frac{x^{3}-1}{x-1} = y^{N}.$$
As Gerhard Paseman commented above, the Diophantine equation
$$ \frac{x^{n} − 1}{x-1} = y^{q} \quad x > 1, \quad y>1, \quad n>2 \quad q \geq 2 \quad \mbox{ (*) }$$
was the subject matter of a couple of papers of T. Nagell from the 1920's. Some twenty-odd years later, W. Ljunggren ...

34

For $n\equiv \pm 4\pmod{9}$ there is no solution to $(1)$. Otherwise, for $n\ge 1$, it is conjectured that there are always solutions, even infinitely many.
There are no analytic results, but heuristics suggest that
given $n$, not $0$ or $\pm 4\pmod{9}$, solutions should occur infinitely often,
asymptotically $c\log(N)$ solutions in $|x|,|y|,|z|<N$, see ...

34

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 is only one point with $z=0$, namely $[1,0,0]$, so you'll always get solutions to your original equeation provided $p>6\sqrt{p}$, i.e., provided $p>36$. ...

answered Jan 17 '19 at 13:17

Joe Silverman

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31

Take any $a,b$ and set $c=a^n+b^n$. Then the triple $(ac^{n-2},bc^{n-2},c^{n-1})$ is a solution of your equation.
Conversely, if $(x,y,z)$ is a solution and $d$ is their gcd, so $(x,y,z)=(ad,bd,cd)$, then you get $d(a^n+b^n)=c^{n-1}$. One of the solutions is presented above (with $d=c^{n-2}$). But there also exist smaller solutions --- they appear as soon ...

31

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 to show that $a_n>0$ for all $n\geq12$.
If $n$ is an odd, let $n=2m+1$, we have $m\geq6$, so by induction hypothesis, there exists $\pi\in S_m$ such that $\...

29

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
$$
ap^x + bq^y = c+ dp^z q^w
$$
has a bounded number of solutions in $(x,y,z,w)$ and that a bound on these could be computed (and the equation solved in practice). This solves (in ...

28

I assume that you're asking if there exists such a solution in $\mathbf{F}_p$ for infinitely many $p$, or for all sufficiently large $p$ under some condition (since for the single equation $x^2 + 1 = 0$ you can't hope for all large $p$ in general).
This can be solved using the Chebotarev Density Theorem and some estimates of Deligne from Weil II (as a ...

27

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 even known that there exists one solution for all $n \not\equiv \pm 4 \pmod{9}$.
Today the smallest number for which the problem is open is $n=114$.
Here is ...

answered Nov 9 '15 at 18:54

Sebastien Palcoux

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26

Though I don't have a full answer to your question, the following remarks may help.
Let's distinguish between (1) explicit examples of systems of Diophantine equations that are known to be undecidable, and (2) a system of Diophantine equations that has the property of being undecidable, whether we know it or not.
Regarding number (1), Jones has written ...

answered Dec 4 '12 at 20:17

Timothy Chow

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26

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 linear change of variables over $\mathbb{Z}_p$
$$x'=ax+bz,\qquad y'=y+(c/a)z,\qquad z'=(1/a)z,$$
we have
$$x'y'-(abc)z'^2=axy+byz+czx.$$
Therefore, the quadratic ...

25

In this generality, the answer is no. The projective curve $X$ given by $2y^2z^2 = x^4 - 17z^4$ over the rationals satisfies the HP, since it has local points everywhere (the affine part $z \neq 0$ is given by $2y'^2=x'^4-17$, which is the famous Reichardt-Lind equation which is known to be everywhere locally, but not globally, soluble) and it has the unique ...

ag.algebraic-geometry nt.number-theory arithmetic-geometry diophantine-equations birational-geometry

25

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 the results of Siegel, Mahler, Lang (see Chapter 5 in Bombieri-Gubler: Heights in Diophantine geometry), your equation also has finitely many solutions.
By a ...

24

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 $\frac{x^n - 1}{x-1}=y^q$, Norsk. Mat. Tidsskrift, 25 (1943), 17 -- 20 ( = Collected Papers of W. Ljunggren edited by P. Ribenboim, Volume 1, #14, p. 363 -- 366)...

24

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 equation
$x^2-3z^2=-3$, so $x+z\sqrt{3}=\pm\sqrt3(2+\sqrt3)^n$.
You can of course replace $x-1$ and $x+1$ by $ax+b$ and $ax-b$ and find
infinitely many ...

23

There are no nontrivial solutions. This follows from Wiles’s proof, and the following observation.
Proposition: If a set of Diophantine equations has a solution in (positive) ordinals using natural sum and product, then it has a solution in (positive) natural numbers.
Proof: Every ordinal can be uniquely written in Cantor normal form
$$\tag{$*$}\alpha=\...

answered Dec 9 '13 at 14:56

Emil Jeřábek

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23

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
$$ E_1: X^3-1=2Y^2,\quad E_2:3X^3-1=2Y^2,\quad E_3:9X^3-1=2Y^2. $$
Next pick out the solutions that have $X$ equal to a power of $3$. These then give the solutions to your equation with $3^n=X$ and $x=Y$. ...

answered Sep 20 '16 at 11:26

Joe Silverman

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22

Fermat never gave a proof, only announced he had one (sounds familiar?). Euler did give a proof, which was flawed, see Franz Lemmermeyer's lecture notes, or see page 4 of David Cox's introduction.
For a discussion why a proof along the lines set out by Fermat is unlikely to work, see this MO posting.
---- trivia ----
As a curiosity, I looked up Fermat's ...

answered Sep 15 '13 at 15:14

Carlo Beenakker

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22

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 stated there)
Suppose first $n$ is even, say $n=2k$. Then the equation is equivalent to $(3^k+1)(3^k-1)=2m^2$. Clearly $\gcd(3^k+1,3^k-1)=2$, so one of these ...

22

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 $4(xy+1)(xy+x+2)=4n^2=(2xy+x+3-z)^2$ for some integer $z=2n-2xy-x-3$, this yields $0<z<x+3$ and rewrites as $z^2-2z(2xy+x+3)+x^2+2x+1=0$. Note that $x$ must divide $z^2-6z+1$, for each $z\...

answered Oct 21 '18 at 22:02

Fedor Petrov

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21

[More a comment than an answer, but a bit long for that]
D. J. Lewis, in Diophantine equations: p-adic methods, pp. 25–75 of Studies in Number Theory, Math. Assoc. Amer. 1969, MR0241359 (39 #2699), writes (p. 28), "As you might expect, there is some artistry in choosing the appropriate modulus."
This was confirmed by Valeriu Şt. Udrescu in the paper, On D....

answered Jun 21 '13 at 6:10

Gerry Myerson

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21

https://sites.google.com/site/tpiezas/010 cites
the following polynomial identity $$(m^3-3^6n^9)^3+(-m^3+3^5mn^6+3^6n^9)^3+
(3^3m^2n^3+3^5mn^6)^3=m(3^2m^2n^2+3^4mn^5+3^6n^8)^3$$ valid for any $n$ and $m$.

21

The short answer to your specific question is no, the resolution of FLT via modularity of elliptic curves does not seem to be helpful in dealing with rational points on higher dimensional varieties. The first two equations you list, $a^5+b^5=c^5+d^5$ and $a^6+b^6=c^6+d^6$, are surfaces of general type in $\mathbb{P}^3$, so conjectures of Bombieri, Lang, ...

answered Nov 30 '13 at 23:13

Joe Silverman

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21

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 fact (at least) two ways of showing this.
$X$ has an automorphism $\tau$ of order 3 defined over $\mathbb Q$ (rotate the projective coordinates). The quotient $X/\...

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