Integer solutions of $2a+2b-ab\gt 0$

$2a+2b-ab>0$ is equivalent to $(a-2)(b-2)<4$, or to $(a-2)(b-2) \le 3$. If $a \ge 3$ and $b \ge 3$, then $(a-2)(b-2) \ge 1$. Thus, $(a-2)(b-2) \in \{1,2,3\}$.

$\bullet$ If $(a-2)(b-2)=1$, then $a-2=b-2=1$, so that $(a,b)=(3,3)$.

$\bullet$ If $(a-2)(b-2)=2$, then $\{a-2,b-2\}=\{1,2\}$, so that $\{a,b\}=\{3,4\}$.

$\bullet$ If $(a-2)(b-2)=3$, then $\{a-2,b-2\}=\{1,3\}$, so that $\{a,b\}=\{3,5\}$.

Therefore, we have the five solutions $(a,b)=(3,3), (3,4), (4,3), (3,5),\:\text{or}\: (5,3)$. $\blacksquare$

Tags:

Inequality

Discrete Mathematics

Diophantine Equations

Elementary Number Theory

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