How can I find the minimum value for $F(x,y,z,w)=x^2+y^2+z^2+w^2+xy+zw-xz-yw-yz$ Let $x,y,z,w$ be integer numbers,and $xw=yz+1$
Find this minimum of the value
$$x^2+y^2+z^2+w^2+xy+zw-xz-yw-yz$$
This is how did it and I would like to know if I made a mistake 
Let $$F(x,y,z,w)=x^2+y^2+z^2+w^2+xy+zw-xz-yw-yz$$
we have that
$$2F(x,y,z,w)=(x-z)^2+(y-w)^2+(x+y)^2+(w+z)^2-2yz$$

let $z=-v ,xw=1-yv,f=x^2+y^2+v^2+w^2+xy-vw+xv-yw+yv=(x-w)^2+2xw+(x-w)(y+v)+(y+v)^2-yv=(x-w)^2+(x-w)(y+v)+(y+v)^2+2-3yv, \\ yv \le \dfrac{(y+v)^2}{4} \implies -3yv \ge -\dfrac{3(y+v)^2}{4} \implies f \ge (x-w)^2+(x-w)(y+v)+\dfrac{(y+v)^2}{4}+2=(x-w+\dfrac{y+v}{2})^2+2 \ge 2 $
when $y=v ,x-w+y=0 \cap xw=1-y^2 ,f $ get min
now we prove the condition is reachable.
$x-w\pm \sqrt{1-xw}=0 \cap xw \le 1 \implies (x-w)^2=1-xw \iff $ 
$\\x^2-xw+w^2-1=0 \\ \tag 1\\x=\dfrac{w \pm \sqrt{4-3w^2}{}}{2}$ 
so there is infinity solution because from (1), we always have $0= x^2+w^2-xw-1 \ge xw-1 \iff xw \le 1$. details of $x-w$ curve  see picture below:

if you want integer ,there are 4 pairs for $x,w$
