Set of OSK 2016 2

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

$x+y+z=t$

$3x + 3y + 3z = 3t$

$3x+3y + x^2 + 2y^2 = 3t$

Let's complete the two squares on the LHS:

$(x^2 + 3x + ( \frac {3}{2} )^2 ) - ( \frac {3}{2} )^2 + (2y^2 + 3y + ( \frac {3}{2 \sqrt {2} } )^2 ) - ( \frac {3}{2 \sqrt {2} } )^2 = 3t$

$(x + \frac {3}{2} )^2 + ( \sqrt{2}y + \frac {3}{2 \sqrt {2} } )^2 - \frac {9}{4} - \frac {9}{8} = 3t$

$(x + \frac {3}{2} )^2 + ( \sqrt{2}y + \frac {3}{2 \sqrt {2} } )^2 - \frac {27}{8} ) = 3t$ ... (i)

Now, each of the squares has a minimum value of 0, hence the minimum value of:

$\text {• 3t is: } - \frac {27}{8}$

$\text {• t is: } - \frac {27}{8} ÷ 3 = \boxed {- \frac {9}{8}}$

This minimum value of t gives us a unique (x, y, z) triplet:

$x = - \frac{3}{2}$

$y = - \frac{4}{3}$

$z = (- \frac{3}{2})^2 + 2 × (- \frac{4}{3} )^2 = \frac{9}{4} + 2 × \frac{16}{9} = \frac{209}{36}$

It is easy to see, that if:

$\text {• } t < - \frac {9}{8} \text { , then we have no solution (since we cannot go lower, than the minimum) }$

$\text {• } t < - \frac {9}{8} \text { , then we have no solution (since we cannot go lower, than the minimum) }$

$\text {• } t > - \frac {9}{8} \text { , then we have an infinite number of solutions }$

(the transformed (separate linear (bijective) transformations) of the x and y coordinates of the (infinite number of) points of a circle):

Let:

$m = x + \frac {3}{2}$

$n = \sqrt{2}y + \frac {3}{2 \sqrt {2} }$

$r^2 = 3t + \frac {27}{8}$

Then (i) can be written as:

$m^2 + n^2 = r^2$

which is an equation of a circle on the (m, n) plane with infinite number of (m,n) solutions for each value of t.

For each (of the infinite number) of (m, n) points, we can get a single (x, y) pair (and from that, a single z)).

Hence, we have an infinite number of (x, y, z) solutions for each value of t > -9/8.