Let's determine the sign of the following numbers:

$$
\sqrt{2}+\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}, \quad \sqrt{6}-\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}
$$
In both expressions, the 2nd and 3rd terms under the square root are so-called "related" expressions, whose product is rational, precisely equal to 1. Therefore, it is expected that if we write our expressions in the form

$$
\sqrt{2}-(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}), \quad \sqrt{6}-(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})
$$

and replace the binomials in parentheses with the square root of their square, we will get a simpler expression. Indeed,

$$
(\sqrt{2+\sqrt{3}} \mp \sqrt{2-\sqrt{3}})^{2}=2+\sqrt{3} \mp 2 \sqrt{1}+2-\sqrt{3}=4 \mp 2
$$

which is 2 or 6, so the value in parentheses is $\sqrt{2}$ or $\sqrt{6}$. Thus, both numbers are equal to 0, and they have no sign.

Tamás Géza (Makó, József A. g. I. o. t.)

Remarks. 1. The binomials in parentheses can be replaced by single terms using the known

$$
\sqrt{A \pm \sqrt{B}}=\sqrt{\frac{A+\sqrt{A^{2}-B}}{2}} \pm \sqrt{\frac{A-\sqrt{A^{2}-B}}{2}}
$$

pair of identities, indeed

$$
\sqrt{A+\sqrt{B}} \mp \sqrt{A-\sqrt{B}}=2 \sqrt{\frac{A \mp \sqrt{A^{2}-B}}{2}}
$$

In our case, $A=2, B=3$, so $A^{2}-B=1$, and thus $\sqrt{2}$ or $\sqrt{6}$ stands on the right side.

Lehel Jenö (Budapest, Apáczai Csere J. gyak. g. II. o. t.)

2. If we write $x$ instead of the "3" under the inner square roots, our expressions become functions of $x$:

$$
y_{1}=\sqrt{2}+\sqrt{2-\sqrt{x}}-\sqrt{2+\sqrt{x}}, \quad y_{2}=\sqrt{6}-\sqrt{2-\sqrt{x}}-\sqrt{2+\sqrt{x}}
$$

Their values at $x=0$ and $x=4$ can be easily obtained. We find that at the former, $y_{1}$ is positive and $y_{2}$ is negative, and at the latter, it is the opposite, i.e., in the graph of both functions, the points corresponding to $x=0$ and $x=4$ are on opposite sides of the $X$-axis. It is natural to think that $y_{1}$ and $y_{2}$ vary continuously, so between $x=0$ and $x=4$ they take on every value between the two found values (and possibly others), including 0. (Intuitively: their graph is a continuous line, and thus it intersects the $X$-axis somewhere.) If we knew that $y_{1}$ and $y_{2}$ take the value 0 only once between $x=0$ and $x=4$, then this value of $x$ would divide the interval from 0 to 4 into two parts, in one of which the function is positive, and in the other, negative, so its sign can be easily determined. By finding which $x$ values satisfy $y_{1}=0$ and $y_{2}=0$ (i.e., solving these two equations for $x$), we find that for both functions, only the point $x=3$ can be considered. Here, the value of our functions is precisely the given two expressions. If, therefore, our assumption about the continuity of the functions is correct, then for $x=3$, $y_{1}=0$ and $y_{2}=0$.

However, the correctness of this reasoning can only be demonstrated with deeper concepts and knowledge than those taught in high school. A more complex form of this line of thought appears in the work of Fazekas Patrik (Mosonmagyaróvár, Kossuth L. g. II. o. t.), who also wrote a variable in place of the "2" number.[^0]


[^0]:    ${ }^{1}$ See, for example, Faragó L.: Collection of Mathematical Circle Problems. Middle School Circle Booklets. (Tankönyvkiadó 1955.) p. 71, problem 80.