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.
Okay. Seen not everyone likes generalizations Pythagorean triples.
Write solutions for degree 3 with three terms.

$X^3+Y^3+Z^3=R^3$

Below are symmetric solutions of this equation. when: $\frac{|X+Y|}{|R-Z|}=\frac{b^2}{a^2}$
Why such solutions found do not know. Probably because beautiful.

$X=3(b^6-7a^6)as^2-9a^4ps-ap^2$

$Y=(6ab^6+21a^7-27b^3a^4)s^2-3(2ab^3-3a^4)ps+ap^2$

$Z=(3b^7+33ba^6-18a^3b^4)s^2+3(4ba^3-b^4)ps+bp^2$

$R=(3b^7+6ba^6-9a^3b^4)s^2+3(2ba^3-b^4)ps+bp^2$

More.

$X=(b^6-7a^6)ap^2+9a^4ps-3as^2$

$Y=(2b^6+9b^3a^3+7a^6)ap^2-3(2ab^3+3a^4)ps+3as^2$

$Z=(3b^4a^3+2ba^6+b^7)p^2-3(2ba^3+b^4)ps+3bs^2$

$R=(6b^4a^3+11ba^6+b^7)p^2-3(4ba^3+b^4)ps+3bs^2$

More.

$X=b(a^6-b^6)p^3+3ba^6p^2s+3ba^6ps^2+b(a^6-b^6)s^3$

$Y=(a^7-ab^6)p^3+3(a^7-a^4b^3-ab^6)p^2s+3(a^7-2a^4b^3)ps^2+(a^7-3a^4b^3+2ab^6)s^3$

$Z=(2ba^6+3a^3b^4+b^7)p^3+3(2ba^6+a^3b^4)p^2s+3(2ba^6-a^3b^4)ps^2+(2ba^6-3a^3b^4+b^7)s^3$

$R=(a^7+3a^4b^3+2ab^6)p^3+3(a^7+2a^4b^3)p^2s+3(a^7+a^4b^3-ab^6)ps^2+(a^7-ab^6)s^3$

More.

$X=sp^6(aj-bt)^2-3t^2j^2p^2s^5+3jt(aj+bt)ps^6-(a^2j^2+abtj+b^2t^2)s^7$

$Y=2sp^6(aj-bt)^2-6jt(aj-bt)p^4s^3+3(a^2j^2-b^2t^2)p^3s^4+3t^2j^2p^2s^5-3tj(aj+bt)ps^6+$
          $+(a^2j^2+abtj+b^2t^2)s^7$

$Z=p^7(aj-bt)^2-3tj(aj-bt)s^2p^5+3bt(aj-bt)s^3p^4+3t^2j^2p^3s^4-6bjt^2p^2s^5-$               $-(a^2j^2-2abtj-2b^2t^2)ps^6$

$R=p^7(aj-bt)^2-3tj(aj-bt)s^2p^5+3aj(aj-bt)s^3p^4+3t^2j^2p^3s^4-6atj^2p^2s^5-$
  $-(b^2t^2-2abtj-2a^2j^2)ps^6$

$a,b,s,p,j,t$ - What are some integers.
Formula course is pointless and unnecessary. But interesting. There is some sort of beauty in them.
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.