A number theory problem by Ilham Saiful Fauzi

Lets rewrite the given expression in this way,

$a^{2}+b^{2}+ab+2(a-b)=4+4+1$

Now, rearrangement and factorisation yields that the expression can also be written into this way, $(a+b+4)(a-2)=(1+b)(1-b)$

From this equation, we can conclude that if $b \geq 1$ ,then, $a \leq 2$ . Now putting $a=1,2$ in the factorised form we easily find that $b=3,1$ respectively.

So, $(a,b)=(2,1),(1,3)$ are the only valid solutions and hence the answer is $\boxed{2}$

As $a,b$ are positive integers we have $a,b \geq 1$ .

Let $X=a+b \geq 2$ and $Y=a-b$ . The condition is equivalent to:

$\dfrac{X^2+Y^2}{2}+\dfrac{X^2-Y^2}{4}+2Y=9 \\ 2X^2+2Y^2+X^2-Y^2+8Y=36 \\ 3X^2+Y^2+8Y=36 \\ 3X^2+Y^2+8Y+16=52 \\ 3X^2+(Y+4)^2=52$

$(Y+4)^2 \geq 0 \implies 3X^2 \leq 52 \implies X \leq 4$

Combining this with $X \geq 2$ gives $X=2,3,4$ . Checking this with the original equation gives the solutions:

$(X,Y)=(3,1),(4,-2)$

$a=\dfrac{X+Y}{2}, b=\dfrac{X-Y}{2}$

$(a,b)=(2,1),(1,3)$

$\color{#20A900}{\boxed{\boxed{\text{2 Solutions}}}}$