Solve $x^3 +1 = 2y^3$ in integers.
(Actually the original question was solve $x^n +1 = 2^{n-2} y^n$ but I can't even solve particular case $n=3$.)
Thanks in advance.
Solve $x^3 +1 = 2y^3$ in integers.
(Actually the original question was solve $x^n +1 = 2^{n-2} y^n$ but I can't even solve particular case $n=3$.)
Thanks in advance.
For a classical proof by Euler: http://books.google.es/books?id=mqI-AAAAYAAJ&hl=es&pg=PA234#v=onepage&q&f=false
Here's a sketch of what I would do.
Instead of considering the equation $x^3 + 1 = 2y^3,$ I would look at the equation
$(-x)^3 + 2y^3 = 1.$
Now what this says is that the pairs $(x,y)$ which solve your equation are in one to one correspondence with elements of the form $-x + \sqrt[3]{2}y$ in $\mathbb{Q}(\sqrt[3]{2})$ of norm $1$ over $\mathbb{Q}$ where $x,y\in \mathbb{Z}.$ So all solutions can be obtained by examining the unit group of the ring of integers of $\mathbb{Q}(\sqrt[3]{2}).$
Note by Dirichlet's unit theorem $\mathcal{O}_{\mathbb{Q}(\sqrt[3]{2})}^{\times} \cong \mathbb{Z}/2 \oplus \mathbb{Z}.$ Furthermore, the free part of this unit group is isomorphic to elements of norm $1.$ Now the problem is 'easy.' Find a primitive unit for this unit group and examine which powers yield elements of the desired form. Good luck!
I've only recently joined Maths SE, so didn't see the question when originally posted.
If $x^3 + 1 = 2y^3$ then $1 (= 1^3), y^3, x^3$ are in arithmetic progression. However, Y. Hellegouarch, in Introduction to the Mathematics of Fermat-Wiles (English translation, Academic Press 2002) states the following on p 342:
Denes Conjecture: Let p be an odd prime. If the three natural non-zero integers $x^p, y^p, z^p$ lie in an arithmetic progression, then x = y = z.
It is stated on p 343 that this conjecture has been proved by Darmon & Merel. The reference is to the following (which I have not seen): Darmon H. and Merel L., Winding quotients and some variants of Fermat's last theorem J. Reine Angew. Math 490 81-100, 1997.
This implies that the above equation has no non-trivial solution in positive integers, and more generally that $x^3 + z^3 = 2y^3$ has no non-trivial solution in positive integers. It would seem however to leave open the possibility of a solution with negative x and y.
Addendum: I realise that the Darmon & Merel result is not necessary to show that $x^3 + z^3 = 2y^3$ has no non-trivial solution in integers. The impossibility is proved in Chapter 2, on p 79 of Sierpinski Elementary Theory of Numbers which I found could be accessed at http://matwbn.icm.edu.pl/kstresc.php?tom=42&wyd=10. This covers the case of negative x and y too.
Mordell, Diophantine Equations, Chapter 23, Theorem 5 (page 203): If $d$ is an integer, $d\gt1$, there is at most one integer solution of $x^3+dy^3=1$ other than $x=1$, $y=0$.
Also, Chapter 24, Theorem 5 (page 220): The equation $x^3+dy^3=1$ ($d\gt1$) has at most one integer solution with $xy\ne0$. This is given by the fundamental unit in the ring when it is a binomial unit, i.e., when the fundamental unit takes the form $x+y\root3\of d$.
Both proofs are fairly long, and take some knowledge of Algebraic Number Theory. Maybe there's some elementary trick I'm not seeing for handling the case $d=2$.