A Tale of $a+b$ and $\dfrac{a}{b}$

-- Method 1

First note that we require $b \ne 0$ . Since $a + b$ must be an integer $n$ we must also have $\dfrac{a}{b} = n \Longrightarrow a = nb$ , and so

$a + b = nb + b = n \Longrightarrow b = n(1 - b) \Longrightarrow n = \dfrac{b}{1 - b} = \dfrac{b - 1 + 1}{1 - b} = -1 + \dfrac{1}{1 - b}$ .

For this to be an integer we then require that $1 - b$ divides $1$ , which only occurs when $1 - b = 1 \Longrightarrow b = 0$ or $1 - b = -1 \Longrightarrow b = 2$ .

But $b$ cannot equal $0$ , so $b = 2, n = -1 + \dfrac{1}{-1} = -2, a = -2 \times 2 = -4$ , and thus there is just $\boxed{1}$ integral solution, namely $(a,b) = (-4,2)$ .

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-- Method 2

First note that we require $b \ne 0$ . Multiplying through by $b$ , the given equation can be written as

$ab + b^{2} = a \Longrightarrow b^{2} + ab - a = 0 \Longrightarrow b = \dfrac{-a \pm \sqrt{a^{2} + 4a}}{2}$ .

Now since $b$ must be an integer we will require that $a^{2} + 4a = n^{2}$ for some positive integer $n$ , in which case

$a^{2} + 4a + 4 = n^{2} + 4 \Longrightarrow (a + 2)^{2} - n^{2} = 4 \Longrightarrow (a + 2 + n)(a + 2 - n) = 4$ .

Next, consider the possible factorizations of $4$ : $4 \times 1, 2 \times 2, -2 \times -2$ and $-4 \times -1$ . Since $a + 2 + n \ge a + 2 - n$ we then assign the greater element (where applicable) of each of these factorizations to $a + 2 + n$ and the other element to $a + 2 - n$ , then add these expressions to eliminate $n$ and subsequently solve for $a$ . In only two cases does this yield an integer value for $a$ :

• (i) $a + 2 + n = 2, a + 2 - n = 2 \Longrightarrow 2(a + 2) = 4 \Longrightarrow a = 0 \Longrightarrow b = 0$ , which we can discard;

• (ii) $a + 2 + n = -2, a + 2 - n = -2 \Longrightarrow 2(a + 2) = -4 \Longrightarrow a = -4 \Longrightarrow n = 0 \Longrightarrow b = \dfrac{-(-4)}{2} = 2$ .

After confirming that $(a,b) = (-4,2)$ satisfies the original equation, we conclude that there is just $\boxed{1}$ integral solution $(a,b)$ .

$b=\frac {a}{b}-a$

$a=\frac {b^2}{1-b}$

If $b$ is an integer, for $a$ to be an integer, $\frac {b^2}{1-b}$ should be an integer

$a=\frac {2b-1}{b-1}-(b-1)$

$a=\frac {1}{b-1} +2 - (b-1)$

Since the last two terms of the above equation are already integers, $\frac {1}{b-1}$ should be an integer, and that occurs only at $b=0,2$ . But since b=0 can't be used, there is only one solution, $(a,b)=(-4,2)$

(a,b) = (-4,2)