Take your time think

We have $x^{2}+y^{2}+1=xyz$ $\frac{x^{2}+y^{2} + 1}{xy} = z$ $\frac{x}{y}+\frac{y}{x}+1=z$

Let above equation be equation one

We see that $\frac{x}{y}$ and $\frac{y}{x}$ any one of them reduces if $x>y$ or $y>x$ therefore no positive value of $z$ will be obtained. Therefore $x=y$ .

This means equation one will be

$\frac{x}{x}+\frac{x}{x}+1=z$ $= \boxed{3}$

i) x=y We have 2x^2+1=x^2*z So x=1, z=3 ☑️

ii) x≠y Suppose there exists (x { 0 },y { 0 }) which are the roots of the equation and x { 0 }+y { 0 }has the minimum value. WLOG, suppose x { 0 }>y { 0 } x { 0 }is the root of equation x^{ 2 }-y { 0 }z { 0 }x+y { 0 }^{ 2 }+1=0 The equation has another root x { 1 } So we have x { 0 }+x { 1 }=y { 0 }z •••••••••• (1) x { 0 }x { 1 }=y { 0 }^{ 2 }+1 ••••••••••(2) From (1), we know x { 1 } is an integer. From (2) 0<x { 1 }=\frac { (y { 0 }^{ 2 }+1) }{ x { 0 } } ≤\frac { (y { 0 }^{ 2 }+1) }{ y { 0 }+1 } <y { 0 } So is another pair of roots of the equation (x { 1 },y { 0 }) which makes x+y smaller. CONFLICTION! So there’s no root pair for x≠y. ∴So z=3

PLEASE HELP ME EDIT THE SOLUTION INTO THE NORMAL MATHEMATICAL EXPRESSION!!!!!!

Case 1): $x=y$ ,
we have $2x^2 + 1 = x^2 z$
so $x=1,z=3 \quad \boxed{\checkmark}$ .

Case 2): $x\ne y$ .
Suppose there exists $x_0, y_0$ which are the roots of the equation and $x_0 + y_0$ has a minimum value.
WLOG, suppose $x_0 >y_0$
$x_0$ is the root of the equation $x^2 - y_0 z_0 x + y_0 ^2 + 1 = 0$ .
The equation has another root $x_1$ . So we have
$x_0 + x_1 = y_0 z \quad \quad (1)$
$x_0x_1 = y_0^2 + 1 \quad \quad (2)$ .
From $(1)$ , we know that $x_1$ is an integer.
From $(2)$ , $0 < x_1 = \frac{y_0^2 + 1}{x_0 } \leq \frac{y_0^2 + 1}{y_0 + 1} ≤ y_ 0$
So y {0}=1 is the only way to take the equal-sign. When y {0}=1, x_{0}=2, z still is 3. So (1,2,3), (2,1,3) are the sets of roots. By Vieta Theorem (5,2,3) and (2,5,3) are also two root sets. And they are the only 5 sets. Except these, there is another pair of roots of the equation $x_1, y_0$ which makes $x+y$ smaller.
CONFLICTION!
So there's no else root pair for $x\ne y$ therefore $z=3$ .