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[SOLVED] Integer solutions of system of equations

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anemone

MHB POTW Director
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Feb 14, 2012
3,802
Find all integer solutions of the system of equations $x+y+z=3$ and $x^3+y^3+z^3=3$.
 

kaliprasad

Well-known member
Mar 31, 2013
1,331
we are given
$x+y+z = 3 \cdots(1)$
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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anemone

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Feb 14, 2012
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Sorry kaliprasad , your answer is not quite right....
 

kaliprasad

Well-known member
Mar 31, 2013
1,331

kaliprasad

Well-known member
Mar 31, 2013
1,331
There was a typo error in first line and I corrected the same. Otherwise I do not find error if any. This may be pointed
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,802
uppose $(x,\,y,\,z)$ is the set of solution.

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 $(4,\,4,\,-5)$.