GeorgeRP wrote: Solve the system of equations in integers: \[ ab + 1 = (c+1)(d+1), \quad cd + 1 = (a-1)(b-1). \] $\iff$ (adding) $a+b=c+d$ and $cd=ab-(a+b)$ So $a,b$ are roots of $x^2-px+q=0$ and $c,d$ are roouts of $x^2-px+q-p$ for some integers $p,q$ So $(a,b,c,d)=\left(\frac{p-u}2,\frac{p+u}2,\frac{p-v}2,\frac{p+v}2\right)$ with $p^2-4q=u^2$ and $p^2-4q+4p=v^2$ (discriminant) $p^2-4q=u^2$ and $p^2-4q+4p=v^2$ is $p=\frac{v^2-u^2}4$ and $q=\frac{p^2-u^2}4$ In order all these numbers are integers we need $u\equiv v\equiv 0\pmod 4$ or $u\equiv v\equiv 2\pmod 4$ Moving $(u,v)\to (2u,2v)$ with new $u,v$ such that $u\equiv v\pmod 2$, we get $p=v^2-u^2$ And solutions $\boxed{(a,b,c,d)=\left(\frac{v^2-u^2}2-u,\frac{v^2-u^2}2+u,\frac{v^2-u^2}2-v,\frac{v^2-u^2}2+v\right)}$ whatever are integers $u\equiv v\pmod 2$

The essential part of IMO 2018 SL N5.