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- Feb 14, 2012

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Find all integer solutions of the system of equations $x+y+z=3$ and $x^3+y^3+z^3=3$.

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- #1

- Feb 14, 2012

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Find all integer solutions of the system of equations $x+y+z=3$ and $x^3+y^3+z^3=3$.

- Mar 31, 2013

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$x+y+z = 3 \cdots(1)$

and

$x^3+y^3+z^3 = 3\cdots(2)$

from (1)

$x+y = 3 - z\cdots(3)$

and from (2)

$x^3+y^3 = 3 - z^3\cdots(4)$

From (3) and (4)

because $x+y$ divides $x^3+y^3$ so $x+y$ divides $3-z^3$ or $3-z$ divides $3-z^3$

so $z-3$ divides $z^3- 3$

as $z-3$ divides $z^3-3^3$ or $z^3 - 27$

so $z-3$ divides $(z^3-3) - (z^3- 27) = 24$

further if we have mod 9 then

$x^3 = 0\, or 1\,or\, -1$

$y^3 = 0\, or 1\,or\, -1$

$z^3 = 0\, or 1\,or\, -1$

as we have $x^3+y^3+z^3 = 3$ so we have $x^3=y^3=z^3 = 1$ mod 9

so $x \equiv y \equiv z \equiv 1\pmod 3$

so we need to take x-3 such that they are 1 mod 3 and factor of 24

they are ${ -8, -2, 1, 4}$

This gives choices for x as $(-5, 1, 4, 7)$

same for y and z and we can checking the sets get $x=y=z=1$

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- #3

- Feb 14, 2012

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Sorry
kaliprasad
, your answer is not quite right....

- Mar 31, 2013

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May be. I would like to know the correct answerSorry kaliprasad , your answer is not quite right....

- Mar 31, 2013

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- Feb 14, 2012

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From the identity $(x+y+z)^3-(x^3+y^3+z^3)=3(x+y)(y+z)(z+x)$, we get $8=(3-z)(3-x)(3-y)$. Since $6=(3-z)+(3-x)+(3-y)$, checking the factorization of 8, we see that the solutions are $(1,\,1,\,1)$, $(-5,\,4,\,4)$, $(4,\,-5,\,4)$ and $(4,\,4,\,-5)$.