1989 OIM Problems/Problem 1

Problem

Find all triples of real numbers that satisfy the system of equations:

$x+y-z=-1$ $x^2-y^2+z^2=1$ $-x^3+y^3+z^3=-1$

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Square the first equation: $x^2+y^2+z^2+2xy-2xz-2yz=1$ Subtract the second equation: $2y^2+2xy-2xz-2yz=0$ This factors as: $2(y+x)(y-z)=0$ This implies that $x=-y$ or $y=z$ .

If $x=-y$ , then substituting into the first equation yields $z=1$ . Substituting all of this into the third equation gives $2y^3=-2$ , so $y=-1$ and $x=1$ . Thus a valid triple is $\boxed{(1,-1,1)}$ .

If $y=z$ , then substituting into the first equation yields $x=-1$ . Substituting all of this into the third equation gives $2y^3=-2$ , so $y=-1$ and $z=-1$ . Thus the other valid triple is $\boxed{(-1,-1,-1)}$ .

Since both triples obviously work, this finishes the proof.

~ eevee9406

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1989 OIM Problems/Problem 1

Problem

Solution

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