Simplify the following expression to its simplest form:

$$
\left(\frac{x+\left(x^{2}-y\right)^{\frac{1}{2}}}{2\left(x+y^{\frac{1}{2}}\right)}\right)^{\frac{1}{2}}+\left(\frac{x-\left(x^{2}-y\right)^{\frac{1}{2}}}{2\left(x+y^{\frac{1}{2}}\right)}\right)^{\frac{1}{2}}
$$
Let's denote the given expression by $z$. First of all, we need to ensure that $z$ makes sense in the realm of real numbers.

To make sense of the $\frac{1}{2}$ exponents, we must assume that their bases are non-negative, that is, that

a) in the denominators, $y \geq 0$

b) in the second term of the numerators, $x^{2}-y \geq 0$,

and we must also examine how the two quotients, or the two numerators and the common denominator, behave in terms of their signs under these two conditions.

Due to condition $b)$, $x^{2} \geq y$, that is,

$$
\text { either } \left.\left.b^{\prime}\right) x \geq y^{\frac{1}{2}}(\geq 0), \quad \text { or } b^{\prime \prime}\right) x<-y^{\frac{1}{2}}(\leq 0)
$$

We cannot allow equality in $\left.b^{\prime \prime}\right)$, because otherwise the denominators would vanish; similarly, in $b^{\prime}$), equality cannot hold in both $\geq$ signs at the same time.

In the case of $b^{\prime}$), the denominators and the first numerator are positive because they have positive terms and no negative terms, and the second numerator is non-negative because

due to $b)$ and $a)$, $x^{2}-y \leq x^{2}$ and thus $\left(x^{2}-y\right)^{\frac{1}{2}} \leq x$;

In the case of $\left.b^{\prime \prime}\right)$, the denominators are negative, both terms of the second numerator are negative, and the first numerator is not positive because $\left.b^{\prime \prime}\right)$ and $\left.a\right)$ imply $x^{2}-y \leq x^{2},\left(x^{2}-y\right)^{\frac{1}{2}} \leq \leq|x|=-x$; therefore, in both sub-cases, both quotients are positive or 0, so $z$ has a meaning for both terms without further assumptions.

We can also state that only one of the quotients can be 0, in the $b^{\prime}$) case the second, and in the $b^{\prime \prime}$) case the first, so $z$ is definitely positive, and thus equal to the (positive) square root of its square. Therefore,

$$
\begin{gathered}
z=\left(z^{2}\right)^{\frac{1}{2}}=\left(\frac{x+\left(x^{2}-y\right)^{\frac{1}{2}}+2\left[x^{2}-\left(x^{2}-y\right)\right]^{\frac{1}{2}}+x-\left(x^{2}-y\right)^{\frac{1}{2}}}{2\left(x+y^{\frac{1}{2}}\right)}\right)^{\frac{1}{2}}= \\
=\left(\frac{2 x+2 y^{\frac{1}{2}}}{2\left(x+y^{\frac{1}{2}}\right)}\right)^{\frac{1}{2}}=1^{\frac{1}{2}}=1
\end{gathered}
$$

Thus, the value of our expression (under the given conditions) is constant. A simpler form is clearly not possible.

András Krámli (Szeged, Radnóti M. gyak. g. I. o. t.)

Remark. Most of the solvers incorrectly gave $\pm 1$ as the simplest form of the expression. As a lesson from their point loss, they should remember well that the $\frac{1}{2}$ exponent of a (non-negative) base $a$ means the non-negative square root, and secondly, we write our solution in the form $\pm 1$ when we are looking for a number whose square is 1. Note the previous sentence from a linguistic point of view as well, from the perspective of precise (rigorous) speech. The use of the definite article "the" indicates that we are talking about a well-defined something - in this case, a number; while the demonstrative pronoun "such" here points to the subordinate clause where the property is expressed that can be true for more than one number, in our case +1 and -1 are those that have it, and no other number does. We only accepted solutions as fully valid that stated that -1 cannot be the value of the expression.