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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 $n = 2$.


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Perhaps take a look at - seems even finding one solution for a single value of $n>2$ can be tricky. – testcase Feb 9 '12 at 19:35
Hi testcase - I've seen that, but it seems like the exact opposite of what I'm looking for. Thanks anyway, though. – DavidDavid Feb 9 '12 at 19:38
For $n = 3$ see . – Qiaochu Yuan Feb 9 '12 at 19:41
For "infinitely many", you might want to add "relatively prime"; otherwise it doesn't add to the question since you can scale any nontrivial tuple by infinitely many factors. – joriki Feb 9 '12 at 19:48
Last I knew this was known for $3$, $4$, $5$, $7$, $8$, and open for $6$, $9$, $10$. – André Nicolas Feb 9 '12 at 20:06

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


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.


















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

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

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

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And how do these compare with the solutions at the links in the comments? – Gerry Myerson Mar 23 '14 at 9:47

These Pythagorean triples can appear in the most unexpected place.

If: $a^2+b^2=c^2$

Then alignment: $N_1^3+N_2^3+N_3^3+N_4^3+N_5^3=N_6^3$







And more:







$a,b,c$ - can be any sign what we want.

And I would like to tell you about this equation:


It turns out the solution of integral complex numbers there. where: $j=\sqrt{-1}$

We make the change:




Then the solutions are of the form:





$p,s$ - what some integers.

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This doesn't actually respond to the question, does it? – Gerry Myerson Mar 20 '14 at 12:15
I do not understand? Formulas little? – individ Mar 20 '14 at 13:20
Formulas little, or formulas big, the guy asked for an $n$th power as a sum of $n$ $n$th powers, but all you have is a cube as a sum of five cubes, and a 5th power as a sum of three 5th powers. What do your formulas have to do with the question? – Gerry Myerson Mar 20 '14 at 21:54
To these formulas do not like? Try it yourself to get at least one, and then scold! – individ Mar 23 '14 at 3:58
You are missing the point. I didn't say I didn't like the formulas, and I didn't say I could do any better. I said, twice, and now a third time, that I didn't see what they had to do with the question DavidDavid asked. You must agree that I am right, or else you would find a way to engage with my comments. – Gerry Myerson Mar 23 '14 at 9:43

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