<s>[INST] A Tribonacci numbers identity for Pythagorean quadruples $a^2+b^2+c^2 =d^2$? We have the known Fibonacci identity for Pythagorean triples,
$$(F_n F_{n+3})^2+(2F_{n+1}F_{n+2})^2 = (F_{2n+3})^2$$
and for Lucas numbers,
$$(L_n L_{n+3})^2+(2L_{n+1}L_{n+2})^2 = (L_{2n+2}+L_{2n+4})^2 = (5F_{2n+3})^2$$
But given the tribonacci numbers,
$$T_n = 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136,\dots$$
where we set $T_0 = 0,\; T_1 = 1$, etc, it seems they obey the analogous Pythagorean quadruples,
$$(-T_{n+3}^2+T_{2n+3}+T_{2n+4})^2+(2\,T_n\,T_{n+1})^2+(2\,T_n\,T_{n+2})^2 = (T_{n-1}^2+\,T_{2n+1}+T_{2n+2})^2$$
Some questions:


*

*This relation was discovered empirically and holds true for $n$ up to the hundreds, but it would be good to know a proof that it is true for all $n$.

*Any tribonacci analogue for higher powers, like (eq.30) $F_{n+1}^3+F_n^3-F_{n-1}^3 = F_{3n}$?

 [/INST] Regarding Question 2, after a session with Mathematica, it turns out that the Fibonacci identity,
$$-F_{n-1}^3+\color{brown}{F_n^3}+F_{n+1}^3 = F_{3n}$$
has a Tribonacci analogue for third (and higher) powers though they are more complicated. The Fibonacci uses only three addends, while the Tribonacci has twelve addends, 
$$\sum_{m=-7}^4 c_m\, T_{n+m}^3= T_{3n}$$ 
and the $c_m$ are constants. Explicitly,
$$a + 30\, \color{brown}{T_n^3} + b  = 412\, T_{3n}$$
where,
$$\begin{aligned}
a &= 6\,T_{n-7}^3+11\,T_{n-6}^3+18\,T_{n-5}^3-48\,T_{n-4}^3-24\,T_{n-3}^3-16\,T_{n-2}^3+24\,T_{n-1}^3\\[2.5mm]
b &=-10\, T_{n+1}^3+5\, T_{n+2}^3+2\, T_{n+3}^3+2\,T_{n+4}^3
\end{aligned}$$
For example, let $n=8$ and $T_8 = 44$, then,
$$6(1^3)+11(1^3)+ 18(2^3) -48(4^3) -24(7^3) -16(13^3)+ 24(24^3)+ 30(44^3) -10(81^3) +5(149^3)+ 2(274^3)+ 2(504^3) = 412\,T_{24} = 412(755476)$$
though it holds true for all integer $n\geq 7$.
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