Distinct natural numbers such that $ab=cd=a+b+c+d-3$ Find the distinct natural numbers $a,b​​,c,d$ who satisfying $ab=cd=a+b+c+d-3$.

Assume $a$ is the largest number among $a,b,c,d$; then $(a-1)b=a+c+d-3$
$$b=(a+c+d-3)/(a-1)<(a+a+a-3)/(a-1)=3$$
Hence, $b=1$ or $b=2$.
If $b=1$, then $a=a+1+c+d-3$. This implies $c+d=2$. Not the ideal pair.
If $b=2$, then $2a=a+2+c+d-3$. This implies $a=c+d-1$.
$$cd=ab=(c+d-1) \times 2$$
Hence, $$(c-2)(d-2)=2$$
Note that $c$ and $d$ are natural numbers.
Hence, $c-2 d-2$ is either $(-1,-2)$ or $(1,2)$. Only $(1,2)$ is the pair we want.
Hence, putting all this together, we get that
$$a = 6, b=2, c=3, d= 4$$
