Solve for diophantine equation
- $x^n + y^n + z^n =1$
- $x^n+y^n+z^n=2$
Is this equation solve-able ?
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2for 1. take $x=1$ and $y=z=0$ and for 2. $x=y=1$ and $z=0$ – Dominic Michaelis Mar 12 '13 at 11:19
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@DominicMichaelis I think he means apart from the trivial solutions. – Ishan Banerjee Mar 12 '13 at 11:21
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Do we need to solve for all of $n, x, y, z$? Or is $n$ fixed? Are they allowed to be negative? – TMM Mar 12 '13 at 11:38
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1@Ishan, it's up to OP to tell us what is meant, not for us to guess. Voting to close until we get some clarification, also some motivation (why these particular equations?), some idea of what OP knows about them, and so on. – Gerry Myerson Mar 12 '13 at 11:50
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1Presumably, $n$ is odd, or these question. When $n$ is even, the equations only have the obvious solutions. – Thomas Andrews Mar 12 '13 at 12:10
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3So: $n$ even iff question odd. – azimut Mar 12 '13 at 12:12
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Forget my past comment: it seems like the OP wants TWO equations and not a system with with two equations...ok. – DonAntonio Mar 12 '13 at 12:17
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4More generally (for $n$ odd), it's unknown if there are any nontrivial solutions to $x^n+y^n=z^n+w^n$, for $n \geq 5$. – Mike Bennett Mar 12 '13 at 15:25
2 Answers
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There are the known identities,
$$(1-9t^3)^3 + (9t^4)^3 + (3t-9t^4)^3 = 1$$
$$(1+6x^3)^3 + (1-6x^3)^3 + (-6x^2)^3 = 2$$
Tito Piezas III
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I didn't find any for 5th or higher odd powers (looked for $-100 \le x,y,z \le 100$). [Note I don't count $(a,-a,1)$ or permutations; looking for three distinct absolute values.]
But for cubes, with $x,y,z \in [-200,200]$ there are three cases of $x^3+y^3+z^3=2$: $$(-161,-54,163),\ (-47,-24,49),\ (-6,-5,7).$$ There are also several cases of $x^3+y^3+z^3=1$, but since 1 is a cube this is a special case of four cubes adding to zero, which there is literature on. The six cases found, again with $x,y,z \in [-200,200]$: $$(-150,73,144),\ (-138,-135,172),\ (-138,-71,144),$$ $$(-103,64,94),\ (-12,9,10),\ (-8,-6,9).$$ My guess is that the equation with 2 on the right might not have much literature on it. And there may be known methods to rule out 1 on the right for odd powers greater than 3.
coffeemath
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