# 3 nonlinear equations in 3 variables - with the same variable raised to different powers

Hoping someone can give me some guidance on solving the following system of 3 nonlinear equations in 3 variables:

\left\{\begin{align*} &x^2+y^2=100\\ &xy+yz=-102\\ &y^2+z^2=117 \end{align*}\right.

Thanks!

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Are you solving over the naturals, the integers, the reals, or complex numbers? – Ross Millikan Sep 5 '12 at 3:47

Hint: Add the first, twice the second, and the third. What happens? Then subtract the first from the third. Separately, add the first and third and subtract twice the second.

Added later: If you are working over the reals, you might as well check for integer solutions first. Note that given one solution, you can multiply all the variables by $-1$ to get another. There aren't many choices for the first equation, and you could just try them. Bingo. Otherwise, I was looking for things easy to factor. The second gives $y(x+z)=-102$, while the third minus the first also has a $x+z$.

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Can you explain how does your hint help? I get the sum of squares, but don't know how to proceed even if we're just working over the reals. – Calvin Lin Oct 24 '13 at 21:28

Observe that $y ≠0$ for finite $x,z$

$z=0\implies y^2=117>100\implies x^2=-17\implies x^2z^2=117(-17)≠(-102)^2$

$x=0\implies y^2=100\implies z^2=17 \implies y^2z^2=1700≠(-102)^2$

So $xyz≠0$

$(3)-(1)\implies z^2-x^2=17\implies z+x=\frac{17}{z-x}$

$(2)\implies y(z+x)=-102\implies y=-\frac{102}{\frac{17}{z-x}}=6(x-z)$

Putting the value of $y$ in $(1),(3)$.

$x^2+36(x-z)^2=100\implies 37x^2-72zx+z^2=100-->(4)$

and $z^2+36(x-z)^2=117\implies 37z^2-72zx+x^2=117-->(5)$

On division of $(4)$ by $(5)$, $\frac{37x^2-72zx+z^2}{37z^2-72zx+x^2}=\frac{100}{117}$

$\implies x^2(37\cdot 117 -100)-72zx(117-100)+z^2(117-37\cdot 100)=0$

$\implies 4229x^2-1224zx-3583z^2=0$

If $x=a\cdot z, 4229a^2-1224a-3583=0-->(6)$ as $xyz≠0$

So, $y=6(x-z)=6(az-z)=6z(a-1)$

So, from (2), $az 6z(a-1)+6z(a-1)z=-102$

or $z^2=\frac{17}{1-a^2}$ where $a$ is a root of $(6)$.

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