Nontrivial integer solutions of $ a^3+b^3=c^3+d^3$

Question

How can we obtain a set of nontrivial solutions of

$$ a^3+b^3=c^3+d^3, $$ for $a,b,c,d\in \mathbb{Z}$ where $(a,b)\neq (c,d)$ and $(a,b)\neq (d,c)$.

Say in the range that $|a|,|b|,|c|,|d| \in [0,30]$, whose absolute values small or eqaul to 30?
How many simple but nontrivial solutions are there around this range?

The simple form the solutions are the better.

Thank you!

p.s. When $d=0$, we know it is impossible due to the Fermat's Last Theorem.

Have a look at the paper Characterizing the Sum of Two Cubes. — Théophile, Jul 15 '18 at 22:45
Could you include your own efforts in trying to answer these two questions? As it is, it looks like an exercise or assigned problem and that you're expecting us to answer it entirely for you. — amWhy, Jul 15 '18 at 22:45
p.s. It is summer. It is not an exercise or assigned problem. It is just something coming out for fun for the summer. — wonderich, Jul 15 '18 at 22:48
One thing I knew is that Plato's number actually works if you want to see my efforts: $$ 3^3 + 4^3 = (-5)^3 + 6^3=91, $$ also formulating a question may be an effort if you contemplate something bigger than the question seems to be. Thanks for all other's useful answers! These are very beneficial (much more than simply complaining or voting down). — wonderich, Jul 16 '18 at 02:46
https://math.stackexchange.com/questions/107570/generalization-of-pythagorean-triples/721310#721310 — individ, Jul 16 '18 at 04:10
A brute-force computational search isn't hard, but also not particularly enlightening. The results of such a search are at https://oeis.org/A051347 — Michael Lugo, Jul 20 '18 at 13:44
This is just a cute coincidence $3^3+4^3+5^3=6^3$ but you knew that, right? — poetasis, Apr 26 '19 at 23:11

nguyen quang do · Answer 1 · 2018-07-21T08:04:37.380

It is possible to give a partial answer by mimicking Kummer's approach to the first case of FLT. As in your other related question, let us introduce $\omega$ = a primitive 3rd root of unity, the quadratic field $\mathbf Q(\omega)$, its ring of integers $\mathbf Z[\omega]$. We intend to solve the diophantine equation $a^3+b^3=c^3+d^3$ under the additional hypothesis that $3\nmid a+b$ (or equivalently $3\nmid c+d$, by Fermat's little theorem). For simplification, let us concentrate on primitive solutions, i.e. suppose that $a, b$ are coprime (and similarly $c,d$ ).To exploit the decomposition $a^3+b^3=(a+b)(a+b\omega)(a+b\omega^2)$, we must recall that $\mathbf Z[\omega]$ is a principal ideal ring, whose group of units coincides with its group of roots of unity, generated by $\pm \omega$.

Lemma: Under our hypotheses, for $m\neq n$ mod 3, the factors $(a+b\omega^m)$ and $(a+b\omega^n)$ are coprime in $\mathbf Z[\omega]$. Proof: Since $\mathbf Z[\omega]$ is principal, let us apply Bézout's theorem by showing that the ideal $J$ generated by our two elements contains $1$. Because $\omega^k$ is a primitive 3rd root of unity whenever $k \neq 0$ mod 3, the quotient $\epsilon_k=\frac {1-\omega^k}{1-\omega}$ is a unit. But $(a+b\omega^m)-(a+b\omega^n)=\omega^m(1-\omega^{n-m})b=\epsilon_{n-m}\omega^m(1-\omega)b$ , and similarly $(a+b\omega^m)\omega^n-(a+b\omega^n)\omega^m=-\omega^m(1-\omega^{n-m})a= - \epsilon_{n-m}\omega^m(1-\omega)a$, hence $(1-\omega)b$ and $ (1-\omega)a \in J$. Since $a,b$ are coprime in $\mathbf Z$, Bézout asserts the existence of $u,v \in \mathbf Z$ s.t. $ua+vb=1$, hence $(1-\omega)ua+(1-\omega)vb=(1-\omega) \in J$, hence also $3\in J$. Moreover, $a+b=(a+b\omega^m)+(1-\omega^m)b=(a+b\omega^ma+b\omega^m)+(1-\omega)\epsilon_mb$, so that $a+b \in J$. Finally, our additional hypothesis implies the existence of $s,t\in \mathbf Z$ s.t. $(a+b)s+3t=1$, hence $1\in J$. OUF

EDIT : Any rational prime $p\neq 3$ is unramified in $\mathbf Q(\omega)$. Let $\delta$= gcd $(a+b,c+d)$ and $N$= the norm of $\mathbf Q (\omega)/\mathbf Q$ . Our additional hypothesis and the lemma then imply that $\frac {a+b}{\delta}=\pm N(c+d\omega)=\pm (c^2+d^2-cd)=\pm ((c+d)^2-3cd)$, and similarly $\frac {c+d}{\delta}=\pm ((a+b)^2-3ab)$. Besides, since $c,d$ are the roots of the quadratic equation $x^2-(c+d)x+cd=0$, the discriminant $\Delta=(c+d)^2-4cd=((c+d)^2-3cd)-cd$ is necessarily a perfect square (=square of an integer), and similarly $((a+b)^2-3ab)-cd$ is a perfect square. Summarizing, $\pm\frac{a+b}{\delta}-cd$ and $\pm\frac{c+d}{\delta}-ab$ must be perfect squares.

EDIT 2: I made a mistake again ! $(a+b\omega)$ and $(c+d\omega)$ could have a common factor. This seems to be a dead end.

Unfortunately my conclusion was too hasty. There is an obvious contradiction with the Plato number. I think I made an error in the derivation of the equality $a+b=\pm(c+d)$. I'll think about it more thoroughly and edit it if possible. — nguyen quang do, Jul 18 '18 at 19:39
I edited my first answer. What I get now is a necessary condition, which can be checked numerically in a limited range of absolute values of the variables. — nguyen quang do, Jul 20 '18 at 12:52
I made a mistake again ! $a+b\omega$ and $c+d\omega$ could have a common factor. This seems to be a dead end. — nguyen quang do, Jul 21 '18 at 08:06

Nontrivial integer solutions of $ a^3+b^3=c^3+d^3$

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