An Interesting Quadratic

Rewrite the equation as $a^{2} - 4a + 4 = y^{2} + 4 \Longrightarrow (a - 2)^{2} = y^{2} + 4 \Longrightarrow (a - 2)^{2} - y^{2} = 4$ .

But the only two perfect squares that differ by $4$ are $4$ and $0$ , which yields $a - 2 = 2 \Longrightarrow a = 4$ and $y = 0$ . So if we are only require $a$ to be positive then the desired answer is $\boxed{1}$ , but if both $a,y$ must be positive then the answer would be $0$ .

$(a,y)=$ $\frac{-1}{2},\frac{3}{2}$

Consider the equation

$n+m=nm$

We may write this as

$n+m=nm\implies nm-n=m\implies n(m-1)=m\implies \frac{m}{m-1}=1+\frac{1}{m-1}=n$

Suppose that $(n, m)$ are positive integers. Thus, $m-1|1\implies m-1=1\implies m=n=2$ .

Now, consider the quadratic equation:

$x^2-(n+m)x+nm=0$

By Vieta's formula, $n, m$ are the roots. As we assume that $nm=n+m=a$ for some positive integer $a$ , we may write

$x^2-ax+a=0$

The quadratic formula yields

$x=n=m=\frac{a\pm\sqrt{a^2-4a}}{2}$

In order for the expression on the right to be an positive integer, as implied, $a^2-4a=b^2$ for some positive integer $b$ . As there is only one possible value for $a$ , as shown above, there is only one positive integer solution to $a^2-4a=b^2$ , which is $(4, 0)$ .