For what values of $t$ is the two-variable operation $x * y=x+y+t \sqrt{x y}$, defined on positive numbers, associative?

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We can only speak of an operation if for all $x, y > 0$, $x * y > 0$ is satisfied, that is,

$$
x + y + t \sqrt{x y} > 0, \quad t > -\left(\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}}\right)
$$

This is precisely satisfied for all positive pairs $(x, y)$ if $t > -2$.

The desired identity $(x * y) * z = x * (y * z)$ clearly holds if $t = 0$. If $t \neq 0$, then it is equivalent to the identity

$$
\sqrt{x y} + \sqrt{x z + y z + t \sqrt{x y} z} = \sqrt{y z} + \sqrt{x y + x z + t \sqrt{y z} x}
$$

Testing with the values $x = 4, y = z = 1$:

$2 + \sqrt{5 + 2 t} = 1 + \sqrt{8 + 4 t}, \quad 1 + \sqrt{5 + 2 t} = \sqrt{8 + 4 t} .6 + 2 t + 2 \sqrt{5 + 2 t} = 8 + 4 t, \quad \sqrt{5 + 2 t} = 1 + t, \quad 5 + 2 t = 1 + 2 t + t^{2}, \quad 4 = t^{2}$.

Thus, apart from $t = 0$, only $t = 2$ is possible. In the latter case, $x * y = x + y + 2 \sqrt{x y} = (\sqrt{x} + \sqrt{y})^{2}$, so indeed $(x * y) * z = (\sqrt{x} + \sqrt{y} + \sqrt{z})^{2} = x * (y * z)$. The requirements of the problem are only satisfied by the values $t = 0$ and 2.

David Papp (Budapest, Szent István Gymnasium, 11th grade)