# UKMT Special (Problem $16$)

Find all real numbers $x, y, z$ which satisfy the simultaneous equations

$x^2 - 4y + 7 = 0$

$y^2 - 6z + 14 = 0$

$z^2 - 2x - 7 = 0$

[UKMT BMO Round $1$ $2012$, Q$3$] Note by Yajat Shamji
8 months, 2 weeks ago

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- 8 months, 2 weeks ago

Step 1) Add all of them and factor the term.

   x^2 - 2x + y^{2} - 4y + z^{2} - 6z + 14 = 0

x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = 0

(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 0


Step 2) If any of the squared term is not 0, it must be a positive real number.

If there is at least one squared term that is not 0, the equation will be false because the right hand sight must be greater than 0.

So, each the square term must be equal to 0.

   (x - 1)^2 = 0    =>    x - 1  = 0    =>    x=1

(y - 2)^2 = 0    =>    y - 2 = 0    =>    y=2

(z - 3)^2 = 0    =>    z - 3 = 0    =>    z=3


(x , y , z)=(1 , 2 , 3) is the only real solution.

I apologize for the bad representation.

- 8 months, 2 weeks ago

Correct method and solution!

- 8 months, 2 weeks ago