Integer solutions of $2a+2b-ab\gt 0$ Let $a\in\mathbb{N}_{\ge 3}$ and $b\in\mathbb{N}_{\ge 3}$. What are the solutions of the Diophantine inequality
$$2a+2b-ab\gt 0?$$
By guessing, I found 5 solutions:
$$\text{1)}\, a=3,\, b=3$$
$$\text{2)}\, a=3,\, b=4$$
$$\text{3)}\, a=4,\, b=3$$
$$\text{4)}\, a=5,\, b=3$$
$$\text{5)}\, a=3,\, b=5.$$
Are these all the solutions? How could I find all the solutions rigorously?
 A: $2a+2b-ab>0$ 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 $(a,b)=(3,3), (3,4), (4,3), (3,5),\:\text{or}\: (5,3)$. $\blacksquare$
A: If $a\geq 6$ (or $b\geq 6$) then $$1={1\over 3} + {2\over 3}\geq {2\over a}+{2\over b}>1$$
A contradiction, so $a,b\leq 5$...
A: Since $a,b$ are playing a symmetrical role we can assume $a\le b$.
Let's set $b=au$ with $u\ge 1$.
The equation becomes $2a+2au-a^2u>0\implies au-2u<2\implies (a-2)<\frac 2u<2$  since $u\ge 1$.
Thus $a<4$ but since $a\ge 3$ then $a=3$.
It remains $6+2b-3b>0\iff b<6$ so $b\in\{3,4,5\}$.
We get solutions $(3,3),(3,4),(3,5)$ and all solutions by adding the swaps $(4,3),(5,3)$.
A: My thinking is: in general $2a +2b$ is only two $a$s and two $b$s but $ab$ is $b$ number of $a$s which can be much more than $2$ if $b$ is large. Likewise $ab$ is $a$ number of $b$s and even if we used two "of those $a$s" to account for $2a$ we still have more than $2b$ is $a$ is large enough. So if $a,b$ are "large enough" then $ab$ will "outweigh" $2a + 2b$. ... So now I need to find the limits of $a$ and $b$ before the "outweighing" occurs.
In detail:
$2a + 2b -ab > 0$
$2a + 2b > ab$
$2b > a(b-2)$
$0 > a(b-2) - 2b$
$0 > a(b-2) - 2(b-2) -4$
$4 > (a-2)(b-2)$
So we may have the following positive integer possibilities (as $a-2 \ge 3-2=1;b-2\ge 3-2=1$ and bothe $a-2,b-2$ are integers):
$(b-2) = 1$ and $a-2 = 1,2,3$
$(b-2) = 2$ and $a-2 = 1$
$(b-2) = 3$ and $a-2 =1$
ANd that's that.
$(a,b) = \{(3,3),(4,3),(5,3), (3,4),(3,5)\}$.
====  Alternatively ====
$2a + 2b - ab > 0$
$2a-ab > -2b$ and $2b -ab > -2a$
$a(2-b) > -2b$ and $b(2-a) > -2a$
$a(b-2) < 2b$ and $b(2-a) < 2b$ and as  $a,b \ge 3$
$3\le a < \frac {2b}{b-2}= \frac {2b-4}{b-2} + \frac 4{b-2} = 2+\frac 4{b-2}$ and $3 \le b < 2+\frac {a-2}$.
So $\frac 4{b-2} > 1$ and $\frac 4{a-2}>2$  so $b-2,a-2 < 4$ and $a,b < 6$.
And if $b = 3,4,5$ we have
$3 \le a < 2 +\frac 4{3,2,1} = 3\frac 13, 4, 6$.
So $a = 3$ if $b > 3$ (and vice versa)
So the five solutions you found are the only ones that exist.
