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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.

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  • $\begingroup$ Do you expect any solutions apart from $x=-1,y=0$? $\endgroup$ Aug 31, 2011 at 19:25
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    $\begingroup$ @Ross: I do: $x = 1, y = 1$. $\endgroup$
    – TMM
    Aug 31, 2011 at 19:33
  • $\begingroup$ I start with what you said but I couldnt continue ...Please write your solution if you dont mind . THANX ! $\endgroup$
    – Mahan
    Aug 31, 2011 at 19:50
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    $\begingroup$ One further hint: If $x$ is even, then $x^3 + 1$ is odd. But $2y^3$ is even. Therefore, the only solutions must be of the form $x = 2k-1$ for some $k$. $\endgroup$
    – JavaMan
    Aug 31, 2011 at 20:18
  • $\begingroup$ When you substitute $x=2k-1$ the only thing that you`ll get is $k(4k^2 -6k+3)$ is perfect square . But what we can do next ? $\endgroup$
    – Mahan
    Aug 31, 2011 at 20:44

4 Answers 4

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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!

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  • $\begingroup$ Thank you !!! Im beginner in number theory so may I ask you to finish your solution ? Id be glad to learn your amazing method in solving Diophantine equation . Thanks again . $\endgroup$
    – Mahan
    Aug 31, 2011 at 22:35
  • $\begingroup$ @jspecter, Is there any good way to find a primitive unit? If not, I would say that it's easier to solve this equation in $\mathbb{Z}[w]$, by replacing $(x,y)$ by $(-x,-y)$, and factorize $x^3 - 1 = (x-1)(x-w)(x-w^2)$, and try to write each factor as a cube or $(1-w)$ times a cube. $\endgroup$
    – user325
    Sep 1, 2011 at 9:57
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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 here. This covers the case of negative x and y too.

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  • $\begingroup$ Adam Bailey; i cannot reach the Sierpinski book; all is a form to be written in unknown Polish language. $\endgroup$
    – user55514
    Jan 20, 2013 at 21:41
  • $\begingroup$ Have replaced the link - to relevant part of Sierpinski's book in English. $\endgroup$ Jan 12, 2022 at 22:00
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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 $d=2$.

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  • $\begingroup$ Thank you !! In fact I had proved that there at most finite solution with the trick you mention , and I got nowhere when thinking about an elementary solution . The author of article (which I take the problem from that) claim that It can be solved with Zsigmondy theorem . $\endgroup$
    – Mahan
    Sep 1, 2011 at 7:22
  • $\begingroup$ You know, it would be helpful if you let us know about this paper (author, title, journal, volume, year, pages). The only Zsigmondy theorem I know of says something about $a^n-1$ always having a prime factor not shared with $a^m-1$ for any $m\lt n$. $\endgroup$ Sep 1, 2011 at 12:56
  • $\begingroup$ Sorry ,Its in Farsi (Persian) so I didnt mention its author or journal ,... $\endgroup$
    – Mahan
    Sep 10, 2011 at 12:36
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For a classical proof by Euler: http://books.google.es/books?id=mqI-AAAAYAAJ&hl=es&pg=PA234#v=onepage&q&f=false

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