Determine all pairs $(a, b)$ of strictly positive integers such that $a b-a-b=12$.
Let's try to factor by $a: a(b-1)-b=12$. We could factor by $b$, but that would break the already done factorization, so we should rather try to factor by $b-1$. We have $a(b-1)-b=$ $a(b-1)-(b-1)-1=(a-1)(b-1)-1=12$ so $(a-1)(b-1)=13$. Since $a$ and $b$ are strictly positive, $a-1$ and $b-1$ are positive. Since 13 is prime, we have $(a-1, b-1)=(1,13)$ or $(13,1)$. Thus $(a, b)=(2,14)$ or $(14,2)$.

It remains to verify the solutions: if $(a, b)=(2,14), a b-a-b=28-2-14=26-14=12$. If $(a, b)=(14,2), a b-a-b=28-14-2=14-2=12$. The solution pairs