Divisors of integers new function

Assume that
$$a*b = n , n \in \Bbb N \tag 1$$

There is unlimited solutions for this equation, because no restrictions for this equation to consider $a$ & $b$ as integers.

So how can we make it forcing $a$ & $b$ always as integers?

Here is the method:

As we know that $sin(\pi \cdot a) = 0$ only when $a$ is integer.
and we also know that $x^2 +y^2=0$ only when $x=0$ & $y=0$ provided that $x$ and $y$ are always real.
So we can say:
$sin(\pi \cdot a) = 0$
$sin(\pi \cdot b) = 0$

Let $x=sin(\pi \cdot a)$ and $y=sin(\pi \cdot b)$.

then $sin^2(\pi \cdot a) +sin^2(\pi \cdot b) = 0 \tag 2$

from equation $(1)$ we get
 $b = \frac na$
substitute it in equation $(2)$

 $sin^2(\pi \cdot a) +sin^2(\pi \cdot \frac na) = 0 \tag 3$

Now we have a equation that $a$ must be a devisor of $n$.

Here is the final formula
$$f(n,a)=sin^2(\pi \cdot a) +sin^2(\pi \cdot \frac na)$$

we can also substitute $sin(\pi \cdot a)$ by Euler reflection formula
$$\Gamma(a) \Gamma(1-a)=\frac {\pi}{sin(\pi \cdot a)}$$
Try to give integer values for $n$ and draw the function, you will see that the roots of this function are the divisors of $n$ that when $f(n,a) = 0$.