Consider the natural numbers x, y, z that simultaneously satisfy the conditions:

Question

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!

Answer 1

If $z=y+1$, we have

$$y(y+1)+1=y^2+y+1$$

We need this to be some perfect square,

$$y^2+y+1=n^2\Leftrightarrow (y+1+n)(y+1-n)=y\Rightarrow (y+1+n)|y$$

But this is impossible, since $(y+1+n)>y$.

If $z=y+2$ we have

$$y(y+2)+1=(y+1)^2\Rightarrow \sqrt{1+x(y+1)}=2023\Rightarrow x(y+1)=2022\cdot 2024$$

Can you proceed from here?

Answer 2

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.

Answer 3

NOTE Condition (ii) is not necessary.

$x\sqrt{yz+1}=2022.2024$, where $x$ is in $[2000,2025]$ and so $\sqrt{yz+1}$ is an integer in $[2021,2046]$.

$233$ is a prime factor of $2022$ and no other integer in $[2000,2046]$ and therefore $\{x,\sqrt{yz+1}\}=\{2022,2024\}$. Hence $yz=2021.2023$ or $2023.2025$.

$17$ is a factor of only $2006$ and $2023$ in $[2000,2025]$. However $59$ is a factor of $2006$ but not of $2021,2023$ or$2025$. Therefore $y$ or $z$ is $2023$.