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I made a note some time ago that I had read in some book that the equation

$$313(x^3+y^3)=t^3$$

has positive integer solutions, but that these are so large that it would be absolutely hopeless to search for them with computers. Unfortunately, I didn't write down where I read this and if you only have the equation, the results Google gives you aren't very helpful. I could only find this so far.

Can someone point me to an article or book where I can read more about this equation? (Preferably something with a proof of the claim above which is accessible even if you're not an expert in number theory.)

Tito Piezas III
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Frunobulax
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    +1.Not relevant to this specific equation, but an interesting (and deeply frustrating :P) fact: such "titanic" Diophantine equations are guaranteed to exist, in a precise sense. For any computable function $f$, there is some $n$ and some Diophantine equation in $k\le n$ variables with all coefficients $\le n$ which has a solution, but no solution in $[0, f(n)]^k$ - that is, we have to look "incomputably" high for solutions to general Diophantine equations. This is a consequence of the solution to Hilbert's Tenth Problem. – Noah Schweber Jan 15 '16 at 13:42
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    Actually, you can find the titanic$x,y$ using computers and some clever shortcuts. A first step is solving $p^3+q^3 = 313^2$ which can then be birationally transformed into the elliptic curve $w^3-432\cdot313^4 = t^2$. If someone would be so kind as to find an initial rational point on this curve, then it is only routine machine manipulation to yield positive integer $x,y$. – Tito Piezas III Jan 15 '16 at 15:29
  • Magma says the curve *does* have a rational point, but since the online version is limited to 120 secs, it does not spell it out. Tease. :( – Tito Piezas III Jan 15 '16 at 15:44
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    @TitoPiezasIII According to this page, an integral solution for $p^3 + q^3 = 313^2 r^3$ is $$\begin{align} p: & {\small +355507307842882624593086325021133856149447336710120844428552934573043094018915289363},\ q: & {\small -354602746692986709129018423204648314355484458881941451025238387384142099383045862152},\ r: & {\small +1517122651849438712721950935044230084378368307868200665761294465082177989014675811} \end{align} $$ – achille hui Jan 15 '16 at 16:19
  • @achillehui: Beautiful! Thanks, I found a positive solution with about 21300 digits. – Tito Piezas III Jan 15 '16 at 17:18
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    @TitoPiezasIII amazing, that's probably too big for the margin. – achille hui Jan 15 '16 at 17:33
  • @achillehui: Hehe, good one. I'm writing the answer now. – Tito Piezas III Jan 15 '16 at 17:41
  • $313$ is a prime congruent to $7$ modulo $9$ and the fact that $313$ is representable as a sum of two cubes, fits well with the following old conjecture of Sylvester still unproved: "Al prime congruent to $4,7$ or $ 8$ modulo $9$ is representable" – Piquito Jan 20 '16 at 00:51
  • Can anyone help me find (with a computer search) nonzero integers $a,b,c,d,e,n$ such that $$a^3-nb^3=c^3-nd^3=e^3$$ where $(a,b,e)$ and $(c,d,e)$ are pairwise coprime and $n^2\ne 1$

    One solution set will do. Thanks.

     – NumThcurious Mar 26 '21 at 02:22

2 Answers2

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The equation $313(x^3+y^3) = t^3$ is equivalent to finding,

$$x^3+y^3=313^2z^3\tag1$$

As Achille Hui points out, Noam Elkies found the solution with $83$-digits,

$$\small x_0 = 355507307842882624593086325021133856149447336710120844428552934573043094018915 289363\\ \small y_0 = -354602746692986709129018423204648314355484458881941451025238387384142099383045 862152 \\ \small z_0 =1517122651849438712721950935044230084378368307868200665761294465082177989014675811$$

Update: Curiously, $3(x_0+y_0) = (3\cdot 7\cdot 8273\cdot 64072783\cdot 125303678787043)^3$.

However, the OP wants them positive. Using the method also discussed in this post, given an initial solution to,

$$ax^3+by^3 = cz^3\tag2$$

then a new one can be derived as,

$$a(-bxy^3-cxz^3)^3 + b(ax^3y+cyz^3)^3 = c(-ax^3z+by^3z)^3\tag3$$

We can use $(3)$ iteratively to find an infinite number of solutions. We have,

$$x_0,y_0,z_0 = +,\,-,\,+\\ x_1,y_1,z_1 = +,\,-,\,-\\ x_2,y_2,z_2 = +,\,-,\,+\\ x_3,y_3,z_3 = +,\,-,\,-\\ x_4,y_4,z_4 = \color{red}{+,\,+,\,+} $$

and so on. So the fourth iteration is all positive. Approximately,

$$x_4 = 1.908757×10^{21389}\\ y_4 = 4.955536×10^{21389}\\ z_4 = 1.095063×10^{21388}$$

They are too long to explicitly write down, but if you have Mathematica, you can retrace the steps taken and see those numbers in all their glory.

Community
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Tito Piezas III
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  • This is really a much surprising result and a beautiful answer – Gottfried Helms Jan 15 '16 at 19:08
  • Is it known that this method gives us the minimal positive solution? OP mentions (citing a source) that it is known that this equation has no solutions which are significantly shorter, but I don't see how your approach would establish that. Don't get me wrong - I nevertheless consider your post to be a good one, I'm just wondering if the other result follows from it. – Wojowu Jan 17 '16 at 21:31
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    @Wojowu: The source refers to titanic numbers as having 1000 or more decimal digits, so this is our lower bound. However, the method described in this post depends on an initial solution. So while it shows an easy way to generate an infinite number of answers, it may also skip others, hence its first positive solution is not necessarily the smallest. As MacLeod points out, there is in fact a smaller one with about $6770$ digits. – Tito Piezas III Jan 19 '16 at 14:14
  • @TitoPiezasIII I see. I thought that titanic numbers are ones with 10000 digits or more. Thanks for your reply. – Wojowu Jan 19 '16 at 14:15
  • @Wojowu: I've just asked MacLeod if positive points $mP$ have a predictable $m$. – Tito Piezas III Jan 19 '16 at 14:18
  • So $313^2$ is representable and $313$ must be also representable according to an old Sylvester's conjecture. It is not frequent that $p$ and $p^2$ be both representable; however if $p$ is congruent to $2$ or $5$ modulo $9$ then $p$ is never representable except the trivial case of torsion $p=2$, but in these cases Oesterlé has proven (with quite specialized means) that both $2p$ and $2p^2$ are representable. – Piquito Jan 20 '16 at 01:19
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The curve $a^3+b^3=N$ is birationally equivalent to the elliptic curve $y^2=x^3-432N^2$ with \begin{equation} a=\frac{36N+y}{6x} \hspace{2cm} b=\frac{36N-y}{6x} \end{equation}

For $N=313^2$, the curve has rank $1$ and generator $G=(x_0,y_0)$ with

$x_0$=426235512202934545020503360093256801131707221692968586587468/216170759226021502298882345008844433022529079715666681

$y_0$=278275087731298331021683520315726613848790652329435004093249928083293904849586928211092140/100506794432879496007544646276171310440319758686599267034949687655666070652158579

which give the solution Noam Elkies tabulated.

The curve has no torsion points so all rational solutions come from points of the form $mP, \, \, m=1,2,\ldots$.

Looking at the above transformations, a positive solution will only arise when $|y| < 36N$ since $x>0$ always.

Experiments show that this first happens when $m=9$ giving a solution with roughly 6770 digits

Allan MacLeod
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  • Since the point $mP$ for $m=9$ is positive, are such $m$ predictable? Are these $m$ in arithmetic progression (for example)? – Tito Piezas III Jan 19 '16 at 14:17
  • The isogeny $P\to 9P$ has degree $81$ Do you know the story $2^{64}$ attached to chess? Numbers involved are greater than $2$ and with exponent greater than $64$!!! – Piquito Jan 20 '16 at 01:08
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    As far as I am aware, one cannot predict where the first positive solution will occur. It all depends on the initial generator and the underlying geometry, – Allan MacLeod Jan 22 '16 at 05:39