Square root equation

We can prove that $x=y=z$ , I put the proof below. $\begin{aligned} \sqrt{x-1}+43&=x \\ \sqrt{x-1}+42&=x-1 \\ a=\sqrt{x-1} \\ a+42&=a^2 \\ (a-7)(a+6)&=0 \\ a&=7 \\ \sqrt{x-1}&=7 \\ x&=\large 50 \end{aligned}$

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Lemma: for strictly increasing function $f(x),h(x)$ . If $\begin{aligned} f(a)&=h(b) \quad\quad(1) \\ f(b)&=h(c) \quad\quad(2) \\ f(c)&=h(a) \quad\quad(3) \end{aligned}$ Then $a=b=c$ .

$Proof:$ Case1: $a,b,c$ are distinct numbers. Let $a>b>c$ . By eqn $(2)$ minus eqn $(3)$ , we got $f(b)-f(c)=h(c)-h(a)$ . As $a>b>c$ , we have $f(b)>f(c)$ , and $h(c)<h(a)$ . Hence, $LHS$ of equation above is positive. However, the $RHS$ is negative, which is contradiction. For $a>c>b, b>a>c, …$ we can prove it similarly.

Case 2: two of $a,b,c$ are same, the other is distinct from them. Without loss of generality, let $a=b\neq c$ . We get $f(a)=f(b)\neq f(c),h(a)=h(b)\neq h(c)$ . From eqn $(2)$ , $f(b)=h(c)\\ f(a)=h(c)$ Substitute $(1)$ inside,we get $h(b)=h(c)$ , which is contradiction.

Hence, the lemma is true. Now it is obvious that $f(a)=\sqrt{a-1}+43, h(a)=a$ are both strictly increasing function. Hence, $a=b=c$ .