Our calculator can only add, subtract, and find the reciprocal. Can we get the result of 1 if the initial number is
a) $\sqrt{19}+92$;
b) $\sqrt[19]{92}$;
c) $\sqrt{19}+\sqrt{92}$ ?

(When calculating, we can use the initial number and every intermediate result multiple times, but no other numbers.)
First, we demonstrate a procedure for two operations that we will frequently use in the following.

1. The multiplication of an initial number $x$ by a rational number.

Let the rational number be $\frac{p}{q}$, where $p$ is an integer, and $q$ is a positive integer. Then

$$
\frac{x}{q}=\frac{1}{q \cdot \frac{1}{x}}=\underbrace{\frac{1}{\frac{1}{x}+\frac{1}{x}+\cdots+\frac{1}{x}}}_{q}, \quad \text { so } \quad \frac{|p| \cdot x}{q}=\underbrace{\frac{x}{q}+\frac{x}{q}+\cdots+\frac{x}{q}}_{|p|}
$$

If $p$ is a negative number, we perform the same with $-x=(x-x-x)$.

2. The calculation of the cube of a number $x$:

$$
x^{3}=\frac{x^{3}-x}{1}+x=\frac{1}{\frac{x^{2}+1-x^{2}}{x^{3}-x}}+x=\frac{1}{\frac{x}{x^{2}-1}-\frac{1}{x}}+x=\frac{1}{\frac{1}{x-\frac{1}{x}}-\frac{1}{x}}+x
$$

Now we provide a procedure to generate any number of the form $\sqrt{x}+y$ (where $x, y$ are rational numbers, $y \neq 0, y^{2} \neq x$). Using the previous methods, we can calculate $(\sqrt{x}+y)^{3}-t$, and then subtract $(\sqrt{x}+y)$ from it by $(x+3 y^{2})$ (which is a rational number):

$$
\begin{gathered}
(\sqrt{x}+y)^{3}-\left(x+3 y^{2}\right)(\sqrt{x}+y)= \\
x \sqrt{x}+3 x y+3 y^{2} \sqrt{x}+y^{3}-x \sqrt{x}-x y-3 y^{2} \sqrt{x}-3 y^{3}=2 x y-2 y^{3}=2 y\left(x-y^{2}\right)
\end{gathered}
$$

This is a non-zero rational number, denoted by $\frac{q}{p}$. We multiply it by $\frac{p}{q}$, and thus we obtain 1.

In the following, we generate 1 from a number of the form $\sqrt[19]{x}$ (b) case). ($x$ is a non-zero rational number.) First, we calculate

$$
(\ldots((\sqrt[19]{x} \underbrace{)^{3}\right)^{3} \ldots\right)^{3}}_{7 \text {-times }}=\sqrt[19]{x^{\left(3^{7}\right)}}=\sqrt[19]{x^{2187}}=\sqrt[19]{x^{115 \cdot 19} \cdot x^{2}}=x^{115} \cdot \sqrt[19]{x^{2}}
$$

Since $x^{115} \neq 0$ is a rational number, multiplying it by $x^{-115}$ gives us $\sqrt[19]{x^{2}}$. Then

$$
\frac{1}{\sqrt[19]{x}-\sqrt[19]{x^{2}}}-\frac{1}{\sqrt[19]{x}+\sqrt[19]{x^{2}}}=\frac{2 \sqrt[19]{x^{2}}}{\sqrt[19]{x^{2}}-\sqrt[19]{x^{4}}}=\frac{2}{1-\sqrt[19]{x^{2}}}
$$

Taking the reciprocal of this and doubling it, we get $\left(1-\sqrt[19]{x^{2}}\right)$, to which we add $\sqrt[19]{x^{2}}$.

In the c) case, we first show that if $x$ and $y$ are rational numbers whose square roots are irrational, then from numbers of the form $a \sqrt{x}+b \sqrt{y}$ (where $a, b$ are rational), we can only obtain numbers of the same form. This is equivalent for addition and subtraction. We examine the reciprocal for $\left(a^{2}+b^{2} \neq 0\right)$:

$$
\begin{aligned}
\frac{1}{a \sqrt{x}+b \sqrt{y}}= & \frac{a \sqrt{x}-b \sqrt{y}}{(a \sqrt{x}+b \sqrt{y})(a \sqrt{x}-b \sqrt{y}}=\frac{a \sqrt{x}-b \sqrt{y}}{a^{2} x-b^{2} y}= \\
& \frac{a}{a^{2} x-b^{2} y} \sqrt{x}+\frac{b}{b^{2} y-a^{2} x} \sqrt{y}
\end{aligned}
$$

If $a \sqrt{x}-b \sqrt{y}=0$, we cannot extend it. However, in this case, $\sqrt{x}=\frac{b}{a} \sqrt{y}$ (if $a=0$, then $b$ would also be zero), so $a \sqrt{x}+b \sqrt{y}=a\left(\frac{b}{a} \sqrt{y}\right)+b \sqrt{y}=2 b \sqrt{y}$, and its reciprocal is $\frac{1}{2 b \sqrt{y}}=\frac{1}{2 b y} \cdot \sqrt{y}$, which is also of the form $c \sqrt{x}+d \sqrt{y}$.

Finally, we show that 1 cannot be expressed in this form; in this case, the answer is negative. Suppose $1=a \sqrt{x}+b \sqrt{y}$, with suitable rational numbers $a$ and $b$. Then $(1-a \sqrt{x})^{2}=b^{2} y$, i.e., $1-2 a \sqrt{x}+a^{2} x=b^{2} y$, $2 a \sqrt{x}=1+a^{2} x-b^{2} y$. Since $1+a^{2} x-b^{2} y$ is rational and $\sqrt{x}$ is irrational, it must be that $a=0$. Thus, $1=b \sqrt{y}$, $\sqrt{y}=\frac{1}{b}$, but $\sqrt{y}$ is irrational. This is a contradiction, so our statement is true.

Based on the work of György András (Budapest, Árpád Gym., II. o. t.)

Remarks. 1. By saying that a given initial number, we mean the following. Not only are its decimal digits on the calculator display, but we also know its "special" representations. For example, in case a), not only the number $\sqrt{x}+y$, but also $x$ and $y$, and even their fraction decompositions, are considered known (not on the calculator, but in our "heads"). This is why we could extract $(\sqrt{x}+y) \cdot\left(x+3 y^{2}\right)$ from $(\sqrt{x}+y)^{3}$.

2. We can also state a more general representation theorem: from the initial number $x$, we can obtain all numbers of the form

$$
y_{k}=\frac{a_{2 k} x^{2 k}+a_{2 k-2} x^{2 k-2}+\ldots+a_{0}}{a_{2 k-1} x^{2 k-1}+a_{2 k-3} x^{2 k-3}+\ldots+a_{1} x}
$$

$$
z_{k}=\frac{a_{2 k} x^{2 k}+a_{2 k-2} x^{2 k-2}+\ldots+a_{0}}{a_{2 k+1} x^{2 k+1}+a_{2 k-1} x^{2 k-1}+\ldots+a_{1} x}
$$

(The $a_{i}$ are rational.) This can be proven by induction on $k$.

Based on the work of Szeredi Tibor (Fazekas M. Fôv. Gyak. Gym., II. o. t.)

Of course, numerous other specific procedures can be provided for cases a) and b).