Solve the following equation

$$
x^{2}+\left(\frac{x}{x+1}\right)^{2}=1
$$
$$
x^{2}+\left(\frac{x}{x+1}\right)^{2}=1
$$

I. solution. Multiplying both sides by $(x+1)^{2}$, we get

$$
x^{2}(x+1)^{2}+x^{2}-(x+1)^{2}=0
$$

The sum of the first two terms on the left side is the difference of two squares:

$$
x^{2}(x+1)^{2}+x^{2}=\left(x^{2}+x+1\right)^{2}-(x+1)^{2}
$$

and thus (2) takes the form

$$
\left(x^{2}+x+1\right)^{2}-2(x+1)^{2}=0
$$

The left side of the equation, as the difference of two squares, can be factored into a product:

$$
\left(x^{2}+x+1\right)^{2}-(\sqrt{2}(x+1))^{2}=\left(x^{2}+(1+\sqrt{2}) x+(1+\sqrt{2})\right)\left(x^{2}+(1-\sqrt{2}) x+(1-\sqrt{2})\right).
$$

The first factor is positive for all (real) $x$, and the roots of the second factor are:

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

We performed reversible steps in the domain of interpretation, so the numbers obtained are also roots of (1).

$$
\text { Csirik János (Szeged, Ságvári E. Gyak. Gimn., III. o. t.) }
$$

II. solution. It is easy to verify that the left side of our equation

$$
x^{2}+\left(\frac{x}{x+1}\right)^{2}=\left(\frac{x^{2}}{x+1}\right)^{2}+2 \frac{x^{2}}{x+1}
$$

so the equation can be written in the form

$$
\left(\left(\frac{x^{2}}{x+1}\right)+2 \frac{x^{2}}{x+1}+1\right)-2=\left(\frac{x^{2}}{x+1}+1\right)^{2}-(\sqrt{2})^{2}=0
$$

Factoring into a product:

$$
\left(\frac{x^{2}}{x+1}+1-\sqrt{2}\right)\left(\frac{x^{2}}{x+1}+1+\sqrt{2}\right)=0
$$

From the first factor, the equation

$$
x^{2}+(1-\sqrt{2}) x+(1-\sqrt{2})=0
$$

is obtained. Its roots are

$$
\frac{\sqrt{2}-1+\sqrt{2 \sqrt{2}-1}}{2} \text { and } \frac{\sqrt{2}-1-\sqrt{2 \sqrt{2}-1}}{2} \text {. }
$$

From the second factor, we do not get real roots.

Tokodi Tamás (Szeged, Ságvári E. Gyak. Gimn., III. o. t.) $\varphi$ for

III. solution. It follows from the written equation that there is an angle $\varphi$ such that $x=\sin \varphi$ and $\frac{x}{x+1}=\cos \varphi$. Then

$$
\frac{\sin \varphi}{1+\sin \varphi}=\cos \varphi
$$

from which

$$
\sin \varphi \cos \varphi=\sin \varphi-\cos \varphi
$$

Squaring and rearranging, we get the quadratic equation for $\sin \varphi \cos \varphi$:

$$
\sin ^{2} \varphi \cos ^{2} \varphi+2 \sin \varphi \cos \varphi-1=0
$$

from which $\sin \varphi \cos \varphi=\sqrt{2}-1$, the absolute value of the other root is greater than 1.

Thus, from the relationship between the roots and coefficients of (3), $\sin \varphi$ and $(-\cos \varphi)$ are the roots of the equation

$$
t^{2}-(\sqrt{2}-1) t-(\sqrt{2}-1)=0
$$

From here, the possible values of $\sin \varphi$, i.e., $x$, are:

$$
\frac{\sqrt{2}-1 \pm \sqrt{2 \sqrt{2}-1}}{2}
$$

It can be verified by substitution that the found values are solutions to the original equation.

Following the solution of Tokodi Tamás

IV. solution. Generally, we solve the equation

$$
x^{2}+\left(\frac{a x}{x+a}\right)^{2}=b^{2}
$$

(In the problem, $a=b=1$).

Introduce the unknown $y=\frac{a x}{x+a}$. Then we arrive at the system of equations

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

For the unknowns $u=x+(-y), v=x \cdot(-y)$, we get the system of equations

$$
\begin{gathered}
u^{2}-2 v=b^{2} \\
a u=-v
\end{gathered}
$$

from which we get the quadratic equation for $u$:

$$
u^{2}+2 a u-b^{2}=0
$$

The roots of this, $u_{1,2}=-a \pm \sqrt{a^{2}+b^{2}}$, are real.

The equations

$$
t^{2}-u_{i} t-a u_{i}=0 \quad(i=1,2)
$$

formed with these have roots $x$ and $-y$.

Accordingly, the possible solutions of the equation are:

$$
x=\frac{u_{i} \pm \sqrt{u_{i}^{2}+4 a u_{i}}}{2}=\frac{-a+e \sqrt{a^{2}+b^{2}}+e \sqrt{-2 a^{2}+b^{2}+2 e a \sqrt{a^{2}+b^{2}}}}{2}
$$

(the value of $e$ is +1 or -1).

Among these, the real values are also roots of the original equation, since if $t=x$ satisfies (5), that is,

$$
x^{2}-u x-a u=0
$$

where $u$ satisfies (4), then from (5)

$$
a x=x^{2}-u x-a u+a x=(x-u)(x+a)
$$

from which

$$
\frac{a x}{x+a}=x-u
$$

Also from (5)

$$
2 a u=2 x^{2}-2 u x
$$

and substituting this into (4), we indeed get

$$
b^{2}=u^{2}+2 a u=u^{2}+x^{2}-2 u x+x^{2}=(x-u)^{2}+x^{2}=\left(\frac{a x}{x+a}\right)^{2}+x^{2}
$$