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.
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.