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The Diophantine equation $x^3+3=2^n$ has the obvious solutions $(-1,1)$,$(1,2)$ and $(5,7)$. I have been wondering if there are any other, but my attempts have been fruitless (I tried factoring it over $\Bbb{Q}(\sqrt[3]{3})$, but I don't know the basic properties of this field, such as what the ring of integers are, the class number etc.). Any help in solving this problem would be greatly appreciated.

Edit: We can actually split this into two, more general equations, namely the elliptic curves $$ x^3+3=y^2 \quad \text{and} \quad x^3+3=2y^2 $$ so this opens up another method for solving it.The first one is, in fact, a special case of the infamous Mordell equation.

Edit 2: Looking at this paper by Tzanakis and De Weger :http://www.math.uoc.gr/~tzanakis/Papers/PracticalSolutionThueEq.pdf I was wondering if we could use the methods explained in Section 3 and extend them to these equations (the methods in the paper, however, require some computational machinery).This could lead to solving a Thue equation, which has only finitely many integral solutions and the general method for solving them is in the paper linked.

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  • $\begingroup$ You missed at least another obvious solution: $(-1,1)$. According to WolframAlpha these are the only possible solutions. $\endgroup$
    – A.P.
    Commented Nov 6, 2015 at 13:39
  • $\begingroup$ Oh wow, that's embarrassing :D $\endgroup$
    – B. S.
    Commented Nov 6, 2015 at 13:40
  • $\begingroup$ I have tried to improve the readability of your question by introducing Tex. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. $\endgroup$
    – A.P.
    Commented Nov 6, 2015 at 13:41
  • $\begingroup$ No worries, everything's fine :) $\endgroup$
    – B. S.
    Commented Nov 6, 2015 at 13:43
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    $\begingroup$ Another possible approach is to consider $x^3+3=2y^3$ and $x^3+3=4y^3$. This implies that $x/y$ is very close to the cube roots of 2 or 4, so close that only finitely many solutions are possible, by Thue's theorem or just results on continued fractions. $\endgroup$
    – Aravind
    Commented Nov 6, 2015 at 14:13

1 Answer 1

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Two cases:

$1)$ $n=2k$. Then $x^3+3=\left(2^k\right)^2$. But $a^3+3=b^2$ has $2$ integral solutions (http://oeis.org/A081119) $(a,b)=(1,\pm 2)$, so $(x,n)=(1,2)$.

$2)$ $n=2k+1$. Then $(2x)^3+24=\left(2^{k+2}\right)^2$. But $a^3+24=b^2$ has $8$ integral solutions (http://oeis.org/A081119, in particular http://oeis.org/A081119/b081119.txt)(I found them with a program, or you can see these tables)
$$(a,b)=(-2,\pm 4),(1,\pm 5),(10,\pm32),(8158,\pm736844),$$ so $(x,n)=(-1,1),(5,7)$.

Mordell Equations $x^3+k=y^2$ for $k\in\Bbb Z_{\neq 0}$ are fully solved when $|k|<10^4$ (were solved in $1998$; see here), so we can always use this method when the numbers aren't very large.

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    $\begingroup$ But what are the methods for solving a Mordell equation in general,i.e. what does a computer program do to solve them? $\endgroup$
    – B. S.
    Commented Nov 6, 2015 at 14:09
  • $\begingroup$ @BogdanSimeonov The oeis links I gave show the complete amount of integral solutions (they're fully solved since $1998$). Once you know there must be exactly $8$ solutions, you can simply find them with a program by testing e.g. $a\in[-10^6,10^6], b\in[0,10^6]$, or see the tables I gave. $\endgroup$
    – user236182
    Commented Nov 6, 2015 at 14:12
  • $\begingroup$ Yes, I understood that, but how are they actually solved is my question, i.e. how have computers solved them for all the k up to 10^4 $\endgroup$
    – B. S.
    Commented Nov 6, 2015 at 14:20
  • $\begingroup$ @BogdanSimeonov It's the three mathematicians in $1998$ who were able to solve it for $|k|<10^4$ in their paper in the link I gave (using efficient algorithms for programs). See here for some parts of the paper, which might give you an idea how they solved it. $\endgroup$
    – user236182
    Commented Nov 6, 2015 at 14:25
  • $\begingroup$ @BogdanSimeonov One approach is this: consider $x^3+3=2y^2$. Now we add $a^3-2$ to both sides for a suitable integer $a>0$ to get: $(x+a)(x^2-ax+a^2)=2(y^2+b)$, now it is sometimes possible to show that there is a prime factor of the L.H.S. such that $-b$ is a quadratic non-residue. A classic example is $y^2=x^3+7$, where adding 1 to both sides works. $\endgroup$
    – Aravind
    Commented Nov 6, 2015 at 14:30

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