Proving a conditional $\forall \,n\in \Bbb Z : \left[P(n) \Rightarrow \lnot Q(n)\right]$ directly I want to give a direct proof of a conditionnal $$\forall\, n\in \Bbb Z : \left[ P(n) \Rightarrow \lnot Q(n) \right]$$ such that $$P(n)= n>2$$ and$$ Q(n)= \exists\ m \in \Bbb Z: (\,m+n=mn \wedge n|m\,)$$ hence, I want to prove that $$\forall\, n\in \Bbb Z : \left[\,n > 2 \Rightarrow  \lnot \exists\ m \in \Bbb Z: (\,m+n=mn \wedge n|m\,) \,\right] $$ is true.
I think that I proved the statement:
I take the converse, so $Q(n) \Rightarrow \lnot P(n)$ and assume $Q(n)$ hence, $n|m$.
It follows that $$m=nk: k\in\Bbb Z $$ Thus, $$k=\frac 1{n-1} \Rightarrow \left[m=\frac n{n-1}\in \Bbb Z \iff n=2\right] \Rightarrow \lnot P(n)$$ 
Therefore, $$\left[ Q(n) \Rightarrow \lnot P(n) \right]\Rightarrow \left[P(n) \Rightarrow \lnot Q(n)\right]$$ is true, by contrapositive.
What would be a direct proof of this statement? Is the contrapositive considered a form of direct proof?

You wish to prove the following statement.
For all integers $n$ such that $n > 2$, there does not exist an integer $m$ such that $m + n = mn$ and $n$ is divisible by $m$. We will prove this directly.
We first note that $mn = n + (m-1)n$. Therefore, $(m-1)n=m$ and $n = \frac {m}{m-1}$. I now cite some previous knowledge you may or may not know of. Two adjacent integers $a$ and $a+1$ do not share any prime factors. Therefore, $n = \frac {m}{m-1}$ is only true when $m-1$ has no prime factors to prevent the division from resulting in a whole number integer $n$. the only numbers with that quality are $-1$ and $1$. Therefore, $m$ can only be one of two values $m = 0$ or $m = 2$. This results in two values for $n$, which are $2$ and $0$.  Note that the only time $m$ exists satisfying the first condition is when the value for $n$ does not satisfy the given statement. Therefore, $m$ does not exist and the statement is true.
Comments/Note: I didn't use the divisibility portion of the statement. That's because I didn't need it. $n$ and $m$ are always equal by the equation and equal to either $2$ or $0$. I can only presume that the mention of divisibility is a red herring meant to throw off the op and make them think more critically about what is and is not useful in a proof.
