Generalization of Pythagorean triples Is it known whether for any natural number $n$, I can find (infinitely many?) nontrivial integer tuples $$(x_0,\ldots,x_n)$$ such that $$x_0^n + \cdots + x_{n-1}^n = x_n^n?$$
Obviously this is true for $n = 2$.
Thanks.

These Pythagorean triples can appear in the most unexpected place.
If:  $a^2+b^2=c^2$
Then alignment: $N_1^3+N_2^3+N_3^3+N_4^3+N_5^3=N_6^3$
$N_1=cp^2-3(a+b)ps+3cs^2$
$N_2=bp^2+3bps-3bs^2$
$N_3=ap^2+3aps-3as^2$
$N_4=-bp^2+3(2c-b)ps+3(3c-3a-2b)s^2$
$N_5=-ap^2+3(2c-a)ps+3(3c-2a-3b)s^2$
$N_6=cp^2+3(2c-a-b)ps+3(4c-3a-3b)s^2$
And more:
$N_1=cp^2-3(a+b)ps+3cs^2$
$N_2=bp^2+3bps-3bs^2$
$N_3=ap^2+3aps-3as^2$
$N_4=(3c+3a+2b)p^2-3(2c+b)ps+3bs^2$
$N_5=(3c+2a+3b)p^2-3(2c+a)ps+3as^2$
$N_6=(4c+3a+3b)p^2-3(2c+a+b)ps+3cs^2$
$a,b,c$ - can be any sign what we want.
And I would like to tell you about this equation:
$X^5+Y^5+Z^5=R^5$
It turns out the solution of integral complex numbers there. where: $j=\sqrt{-1}$
We make the change:
$a=p^2-2ps-s^2$
$b=p^2+2ps-s^2$
$c=p^2+s^2$
Then the solutions are of the form:
$X=b+jc$
$Y=-b+jc$
$Z=a-jc$
$R=a+jc$
$p,s$ - what some integers.
