Distinct natural numbers such that $ab=cd=a+b+c+d-3$

Question

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

Answer 1

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 $$a = 6, b=2, c=3, d= 4$$

Answer 2

Find natural numbers $(a,b)$ such that:

$c$ = $\dfrac{1}{2}(a b-\sqrt{(a b-a-b+3)^2-4 a b)}-a-b+3 )$
and $d$ = $\dfrac{1}{2}$ $(a b+\sqrt{(a b-a-b+3)^2-4 a b)}-a-b+3)$

are natural numbers.