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) $F_{n+1}^3+F_n^3-F_{n-1}^3 = F_{3n}$?

 A: An "arithmetic" subsequence of a linear recurrent sequence is again a linear recurrent sequence, of the same order.
A product of linear recurrent sequences (of order $a$ and $b$) is again a linear recurrent sequence, of order at most $ab$.
A sum of linear recurrent sequences (of order $a$ and $b$) is again a linear recurrent sequence, of order at most $a+b$.
Hence, the difference $(-T_{n+3}^2+T_{2n+3}+T_{2n+4})^2+(2\,T_n\,T_{n+1})^2+(2\,T_n\,T_{n+2})^2 - (T_{n-1}^2+\,T_{2n+1}+T_{2n+2})^2$ is a linear recurrent sequence, of order at most $702$. So in order to prove that it is the zero sequence, you can just compute the first $702$ values.
You can cut down on the order (because $T_{1400}$ is not really a small number, right ?) by keeping track at each step of the roots of the characteristic polynomial of the recurrence.
A: 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 $n\geq 7$.
A: Here is another way to express 
$x^2+y^2+z^2=g^2$
x=a+pk
y=b+pk
z=c+pk
g=a+b+c+2pk 
and 
$a == (-2 b c - 2 b pk - 2 c pk - pk^2)  /  (2 (b + c + pk))$
Its handy because 
a=g-z-y
b=g-z-x
c=z-x-y
pk=x+y+z-g   
So you can map your equation into this which is a general proof.

2 more useful equations that satisfy 
$x^2+y^2+z^2=g^2$
The equation exploiting the reversibility of 2nd order equations like this is slightly different so I am using aa instead of a and  k instead of pk etc to avoid confusion
$2k=x_2+y_2+z_2-g_2$
$x_2 = aa + k$
$y_2 = bb + k$
$z_2 = cc + k$
$g_2 = aa + bb + cc + k$    
$aa = (-bb cc + k^2)/(bb + cc)$
$x_3 = k-aa$
$y_3 = k-bb$
$z_3 = k-cc$
$g_3 = k-aa-bb-cc$    
so using your original equations
$x_2$=$(−(T_n_+_3)^2+(T_n_+_3)+(T_n_+_4))$
$y_2$ =$2(T_n)(T_n_+_1)$
$z_2$ =$2(T_n)(T_n_+_2)$
$g_2$ =$(T_n_−_1)^2+(T_2_n_+_1)+(T_2_n_+_2)$
then 
$2k=x_2+y_2+z_2-g_2$
$aa = k + g_2 - z_2 - y_2$
$bb = k + g_2 - z_2 - x_2$
$cc = k + g_2 - x_2 - y_2$   
Then prove that you meet the condition
$aa = (-bb cc + k^2)/(bb + cc)$
which will prove that both equations are true and one of them will be exactly your starting equation and the other will be different and possibly trivial.  If you can show that your equation meets the condition for all n then you are done.
If you cannot there is a second possibility to show that your equation is true for a specific n and that the 2nd equation is your equation for a n+t value and as such your single instance proves a 2nd unique answer that proves a 3rd etc.  You will need to deal with avoiding trivial solutions if you go that route.  This could prove infinite solutions but wouldn't necessarily give ALL solutions.

If you map your equation into $x_3,y_3,z_3,g_3$ instead of $x_2,y_2,z_2,g_2$ you should avoid all trivial solutions because you are ascending in value and will never reach zero.

To restate simply using ascending values which you need for inductive proof.
Your statement $a^2+b^2+c^2=d^2$  stated as you did in Tribonacci numbers depends on 
$a_2^2+b_2^2+c_2^2=d_2^2$
Formed from the absolute value of your original Tribonnaci numbers 
EDITED to fix math error:
Corrected to -a+2k, -b+2k, -c+2k, -d+2k
$a_2= - b - c + d$
$b_2= - a - c + d$
$c_2= - a - b +  d$
$d_2= a + b + c - 2 d$   
Also expressed exclusively in Tribonacci numbers
If you can show that increasing your n value by 1 will produce the same values then its a complete proof of your conjecture.
It is possible this is just another statement and doesn't help you at all but this is a way for you to generate an infinite number of equations any 1 of which can prove your statement provided you can express it with your original equation with n increased by 1.
