242. Solve the equation

$$
1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdot}} \cdot \cdot_{1+\frac{1}{x}}}=x
$$

(in the expression on the left, the fraction sign repeats $n$ times).
242. We will sequentially simplify the fraction on the left:

$$
\begin{array}{rl}
1+\frac{1}{x}=\frac{x+1}{x} ; 1+\frac{1}{\frac{x+1}{x}}=1+\frac{x}{x+1} & =\frac{2 x+1}{x+1} \\
1 & 1+\frac{1}{\frac{2 x+1}{x+1}}=1+\frac{x+1}{2 x+1}=\frac{3 x+2}{2 x+1} ; \ldots
\end{array}
$$

Ultimately, we will obtain an equation of the form

$$
\frac{a x+b}{c x+d}=x
$$

where \(a, b, c\), and \(d\) are unknown integers. This equation is equivalent to the quadratic equation \(x(c x+d)=a x+b\); from this, it follows that our original equation can have no more than two distinct roots (it cannot be an identity, as otherwise it would be satisfied by any value of \(x\), and \(x=0\) clearly does not satisfy the equation).

However, it is not difficult to find two roots of our equation. Indeed, suppose that \(x\) is such that

$$
1+\frac{1}{x}=x
$$

In this case, sequentially simplifying our fraction "from the end," we will obtain

$$
1+\frac{1}{x}=x ; 1+\frac{1}{x}=x ; 1+\frac{1}{x}=x ; \ldots
$$

and ultimately arrive at the identity

$$
x=x
$$

Thus, we see that the roots of the equation \(1+\frac{1}{x}=x\), or \(x^{2}-x-1=0\), i.e.,

$$
x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2}
$$

satisfy the given equation. This equation cannot have any other roots.

Note. Let us outline another method for solving the given equation (compare with the note to the solution of problem 241). Substitute into the left side of the equation the expression for \(x\) as a multi-story fraction given by the equation itself. In this case, we will obtain the same equation, where, however, the fraction sign in the multi-story fraction on the left will be repeated \(2n\) times. Continuing this process, we will arrive at equations where the fraction on the left will have more and more "floors." On this basis, we can write:

$$
\begin{aligned}
x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+.}}}} & =1+\lim _{N \rightarrow \infty} \frac{1}{1+\frac{1}{1+\frac{1}{1+.}}} \cdot++1+\frac{1}{1}
\end{aligned}
$$

## The fraction sign is repeated \(N\) times

where on the left stands an infinite expression. From this, we obtain:

$$
\left.x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+!}}}}=1+\frac{1}{\left[1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdot}}}\right.}=1+\frac{1}{x}, \text { (}^{* *}\right)
$$

i.e., the same quadratic equation for \(x\) that appeared in the solution provided above. This solution of the original equation immediately shows that the roots do not depend on \(n\).

The reasoning presented in this form cannot be considered a solution to the problem, as the existence of the limit (*) and the equality of \(x\) to this limit have not been proven, and the transformations (**) have not been justified. However, this reasoning can be made entirely rigorous.