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Consider the equation $$ X_1^3+X_2^3+X_3^3 = Y_1^3+Y_2^3+Y_3^3. \tag{$\star$} $$

Does this have a complete solution (a.k.a. parameterization) in integers? If not, does it have a complete solution in rational numbers?

I know the [integral and/or rational] solutions to other related equations — such as the 3.2.2 equation $$ X_1^3+X_2^3 = Y_1^3+Y_2^3 $$ and the 3.1.3 equation $$ X_1^3 = Y_1^3+Y_2^3+Y_3^3. $$ (see, for example, Choudhry’s paper) — but I have yet to find any paper giving the complete rational or integral solution to ($\star$).

Kieren MacMillan
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1 Answers1

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Update:

Yes, there is a complete rational solution. Since we are dealing with a homogeneous equation, it is best to consider rational points on,

$$x_1^2+x_2^2 =1\tag1$$

$$x_1^3+x_2^3+x_3^3 =1\tag2$$

$$x_1^3+x_2^3+x_3^3+x_4^3+x_5^3 =1\tag3$$

to reduce the number of variables. The complete rational solution to $(1)$ and $(2)$ are well-known. It turns out the complete rational solution for $(3)$ is,

$$\color{blue}{\big(a + (a + b)\gamma\big)^3 + (a \gamma - b)^3 + \big(c + (c + d)\gamma\big)^3 + (c \gamma - d)^3 + \gamma^3 = 1}\tag4$$

where,

$$\gamma=\frac{-a^3 + b^3 - c^3 + d^3 + 1}{a^3 + (a+b)^3+c^3+(c+d)^3+1}\tag5$$

and can be found by generalizing the approach used by Choudhry in the paper cited by the OP.

Proof:

Equating the five terms of $(3)$ and $(4)$, one can easily solve for the five unknowns $a,b,c,d,\gamma$ as,

$$a = \frac{x_1+x_2x_5}{\beta},\quad b = \frac{-x_2+(x_1-x_2)x_5}{\beta}$$ $$c = \frac{x_3+x_4x_5}{\beta},\quad d = \frac{-x_4+(x_3-x_4)x_5}{\beta}$$ $$\gamma = x_5, \quad\quad\beta = x_5^2+x_5+1$$

Substituting these into $(5)$, the equality holds only if the $x_i$ satisfies $x_1^3+x_2^3+x_3^3+x_4^3+x_5^3 =1$, thus proving the assertion.

Example:

The smallest solution to $(3)$ is,

$$\big(\tfrac{1}{4})^3+\big(\tfrac{1}{4})^3+\big(\tfrac{2}{4})^3+\big(\tfrac{3}{4})^3+\big(\tfrac{3}{4})^3=1\tag6$$

Equating terms with $(4)$ one finds $a,b,c,d,\gamma$ as,

$$a=\tfrac{7}{37},\;b=\tfrac{-4}{37},\;c=\tfrac{17}{37},\;d=\tfrac{-15}{37},\;\gamma=\tfrac{3}{4}$$

If there is a zero term,

$$0+\big(\tfrac{1}{7})^3+\big(\tfrac{1}{7})^3+\big(\tfrac{5}{7})^3+\big(\tfrac{6}{7})^3=1\tag7$$

then,

$$a=\tfrac{6}{127},\;b=\tfrac{-13}{127},\;c=\tfrac{37}{127},\;d=\tfrac{-59}{127},\;\gamma=\tfrac{6}{7}$$

and so on.

Tito Piezas III
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  • What's the approach for such a case? $$x^3+y^3+z^3=a^3+b^3$$ – individ Sep 21 '16 at 17:36
  • @individ: The same general identity covers it, but $a,b,c,d,e$ has to be chosen so one of the $x_i$ vanishes. – Tito Piezas III Sep 22 '16 at 07:14
  • You, sir, are my hero. =) – Kieren MacMillan Sep 22 '16 at 10:49
  • @KierenMacMillan: Pleasure to be of service. =) – Tito Piezas III Sep 22 '16 at 12:00
  • Do you see any reason why this method can't be generalized to 9 cubes, e.g., $$ x_1^3+x_2^3+x_3^3+x_4^3+x_5^3+x_6^3+x_7^3+x_8^3=1$$ ?? – Kieren MacMillan Sep 22 '16 at 13:29
  • @KierenMacMillan: It can be generalized to the sum of $2n$ cubes. I've edited the formula to highlight the variable $\gamma$. For $n=8$ cubes, you would need to add two terms using variables $e,f$ easily extrapolated from the formulas above. For $n=10$ cubes, add four terms using $e,f,g,h$ and so on. For $n=9$, it will be just a special case of the $n=10$ where one term vanishes, similar to individ's question. – Tito Piezas III Sep 22 '16 at 13:46
  • Thanks! Last question: If we write $a=a_1/a_2$, etc., in lowest terms, with $a_1,a_2,\dots$ integers, and then clear denominators, does this give us the complete solution in integers? – Kieren MacMillan Sep 22 '16 at 16:35
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    @KierenMacMillan I believe it does. You can scale $(4)$ so the last addend is $e^3$, not $1$, if you want $5+1$ variables (like Choudhry's $3+1$). And corresponding changes to $(3)$ also. – Tito Piezas III Sep 23 '16 at 03:05