# 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.

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Perhaps take a look at en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture - 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 math.fau.edu/richman/cubes.htm . – 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.

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

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.

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

$N_1=cp^2-3(a+b)ps+3cs^2$

$N_2=bp^2+3bps-3bs^2$

$N_3=ap^2+3aps-3as^2$

$N_4=-bp^2+3(2c-b)ps+3(3c-3a-2b)s^2$

$N_5=-ap^2+3(2c-a)ps+3(3c-2a-3b)s^2$

$N_6=cp^2+3(2c-a-b)ps+3(4c-3a-3b)s^2$

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.

$X^5+Y^5+Z^5=R^5$

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.

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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