Generalization of Pythagorean triples

Question

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.

Answer 1

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.