Consider the natural numbers $x, y, z$ that simultaneously satisfy the conditions:
i) $x,y,z \in \{2000, 2001, 2002, \ldots, 2025\}$:
ii) $|y-z| \le 2$
iii) $\sqrt{1+x\sqrt{yz+1}}=2023$
Show that one of the numbers $x, y, z$ is equal to $2023$.
MY IDEAS
Let $x \le y \le z$
$z-y$ must be $\le 2$ then $z-y$ can be $0, 1$ or $2$.
If $z-y=0$, then $z=y$
The equality will become
$\sqrt{1+x\sqrt{{z}^{2}+1}}=2023$
${z}^{2}+1$ is irrational and will make all the equation irrational. But 2023 isn't irrational, which makes this case impossible.
I tried doing the same for $z-y=1$ and $z-y=2$ but got nowhere. Like I don't know what to do.
Hope one of you can help me. Any ideas are welcome! Thank you!
Best Answer
HINT.- It is necessary that $\sqrt{yz+1}$ be square so we have to look at the functions $$f(t)=\sqrt{t^2+1}\\g(t)=\sqrt{t(t+1)+1}\\h(t)=\sqrt{t(t+2)+1}$$ It is clear that with $h(t)=\sqrt{(t+1)^2}$ we are done.