Solve the equation

$$
x^{2}-4 x+3=\sqrt{1+\sqrt{x-1}}
$$
I. Solution 1: The number under the square root is non-negative: $x \geq 1$. Since the right side is positive in this case, the left side must also be positive: $x^{2}-4 x+3=(x-1)(x-3)>0$. Since $x>3$. Rearranging the equation:

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

The function $f:(3 ;+\infty) \rightarrow(2 ;+\infty), f(x)=(x-2)^{2}+1$ is strictly increasing and bijective on its domain, hence invertible. Its inverse function is $f^{-1}:(2 ;+\infty) \rightarrow(3 ;+\infty), f^{-1}(x)=\sqrt{x-1}+2$. Taking the value of both sides at $f^{-1}(x)=\sqrt{x-1}+2$:

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

The right side is precisely the right side of (1). Based on this, the equation is:

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

which is true if and only if $f^{-1}(x)=f(f(x))$, or $x=f(f(f(x)))$.

Since the function $f$ is strictly increasing, if $f(x)>x$ then $f(f(x))>f(x)>x$, from which $f(f(f(x)))>f(f(x))>f(x)>x$ follows. Similarly, if $f(x)<x$ then $f(f(x))<f(x)<x$, and $f(f(f(x)))<f(f(x))<f(x)<x$. Therefore, the only solution is $f(x)=x$, which gives $x=\frac{5+\sqrt{5}}{2}$, satisfying the condition $x>3$. Substituting back into the original equation, both sides yield $\frac{1+\sqrt{5}}{2}$, so the solution is valid.

II. Solution. Since $x=1$ is not a solution, let $a=\sqrt{x-1}>0$. Then $x^{2}-4 x+3=(x-1)(x-3)=a^{2}\left(a^{2}-2\right)$ and $\sqrt{1+\sqrt{x-1}}=\sqrt{1+a}$. From this, $a^{2}\left(a^{2}-2\right)=\sqrt{1+a}$, squaring both sides gives $a^{4}\left(a^{4}-4 a^{2}+4\right)=1+a$. Rearranging, $a^{8}-4 a^{6}+4 a^{4}-a-1=0$.

Let's try to factor the left side:

$$
\begin{aligned}
a^{8}-4 a^{6}+4 a^{4}-a-1= & a^{8}-a^{7}-a^{6}+a^{7}-a^{6}-a^{5}-2 a^{6}+2 a^{5}+2 a^{4}- \\
& -a^{5}+a^{4}+a^{3}+a^{4}-a^{3}-a^{2}+a^{2}-a-1= \\
= & \left(a^{2}-a-1\right)\left(a^{6}+a^{5}-2 a^{4}-a^{3}+a^{2}+1\right)
\end{aligned}
$$

This is zero if and only if one of the factors is zero.

Examining the second factor:

$$
\begin{aligned}
a^{6}+a^{5}-2 a^{4}-a^{3}+a^{2}+1 & =a^{6}-2 a^{3}+1+a^{5}-2 a^{4}+a^{3}+a^{2}= \\
& =\left(a^{3}-1\right)^{2}+a^{3}(a-1)^{2}+a^{2}>0
\end{aligned}
$$

since the first two terms are non-negative and the third is positive.

If the first factor is zero: $a^{2}-a-1=0$, then $a_{1,2}=\frac{1 \pm \sqrt{5}}{2}$. Since $a>0$, only the positive solution is valid: $a=\frac{1+\sqrt{5}}{2}$. Since $x=a^{2}+1$, we get

$$
x=1+\frac{6+2 \sqrt{5}}{4}=\frac{10+2 \sqrt{5}}{4}=\frac{5+\sqrt{5}}{2}
$$

Substituting this back, it indeed gives a valid solution.

Note. Anyone who factored an 8th-degree polynomial without justification could receive a maximum of 3 points.

III. Solution. From the previous solutions, $x \geq 3$ and the equation $(x-1)(x-3)=\sqrt{1+\sqrt{x-1}}$.

Let $a=x-1 \geq 2$, then

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

or $(a-1)^{2}=1+\sqrt{1+\sqrt{a}}$.[^0]

The graph of the function $y=(x-1)^{2}$ for $x \geq 2$ is a strictly increasing part of the corresponding parabola, which is convex throughout. The function $y=1+\sqrt{x}$ is a root function, whose behavior does not fundamentally change with an additional root (no inflection point is created, it remains concave throughout). Thus, the two graphs intersect at one point.

![](https://cdn.mathpix.com/cropped/2024_05_02_f7682f35c97b1aecb35cg-2.jpg?height=588&width=569&top_left_y=340&top_left_x=775)

Since $a>0$, we can take the square root of both sides: $a-1=\sqrt{1+\sqrt{1+\sqrt{a}}}$, from which

$$
a=1+\sqrt{1+\sqrt{1+\sqrt{a}}}
$$

If we substitute $a$ on the right side with its value, and repeat this process infinitely, we get an infinite sequence: $1+\sqrt{1+\sqrt{1+\sqrt{\cdots}}}$. We conjecture that the value of $a$ in the equation is this sequence, so $a$ can be substituted anywhere within the sequence, thus:

$$
a=1+\sqrt{1+\sqrt{1+\sqrt{\cdots}}}=1+\sqrt{a}, \quad \text { from which } \quad a-1=\sqrt{a}
$$

From this, $x-2=\sqrt{x-1}$, squaring both sides gives $x^{2}-4 x+4=x-1$, rearranging

$$
x^{2}-5 x+5=0
$$

The only solution that satisfies the condition is $x=2.5+\sqrt{1.25}$. Since there are no other roots, this is the only solution.

Note. Since $a>0$, taking the square root of both sides of (2) gives

$$
\sqrt{a}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{a}}}}
$$

with $f(y)=\sqrt{1+y}$, we get the form $y=f(f(f(y)))$ from the first solution.


[^0]:    ${ }^{1}$ It is worth studying Gábor Ábrahám's article "On equations of the type $f^{-1}(x)=f(x)$, or the responsibility of examiners and other interesting things" (KöMaL, December 2010, and www.komal.hu/cikkek/abraham/abraham.h.shtml).