Mistakes give rise to Problems- 8

Observe that given condition can be transformed to a useful form as follows:-

$a+b=ab$

$a-ab+b=0$

$a(1-b)+b=0$

$a(1-b)+b-1=-1$

$a(1-b)-1(1-b)=-1$

$(1-b)(a-1)=-1$

$(a-1)(b-1)=1$

From this you can conclude that because both $a$ and $b$ are integers, $(a-1) =\pm 1$ and $(b-1)=\pm 1$ accordingly.

$\color{#D61F06}{\textbf{Thus the only solutions are}}$

$(a-1)=(b-1)=1$ and $(a-1)=(b-1)=-1$

This gives $\boxed{2}$ pairs $(a,b)$ , namely $(0,0)$ and $(2,2)$

This is the most pure and simple approach:

$ab=a+b$

Dividing by $a$ we get

$b=1+\dfrac{b}{a}.$

This suggests that $b \geq a$ , since the RHS must be an integer. But doing likewise with $b$ , we get

$a=1+\dfrac{a}{b}$

which suggests $a \geq b$ . For both conditions to be satisfied, clearly $a=b$ . Now, substituting, we get

$a^2=a+a$

which has solutions $a=b=0,2$ . Thus there are $\boxed{2}$ solutions.

For me, I thought of 2 simple equations: 2+2=2(2) and 0+0=0(0)