Discrete Mathematics: $mn + 2m + 2n + 2 = n$ proof of uniqueness of $m$, $\forall n \in \mathbb{Z}$ Prove: There exists a unique integer $m$ such that for every integer $n$:
$$mn + 2m + 2n + 2 = n$$
However I am not sure if my proof is correct. How do I prove uniqueness of $m$?
I prove it by solving the equation for $m$. For every $n$ there is a unique $m$ expressed in terms on $n$ such that $mn + 2m + 2n + 2 = n$.
$$m=\frac{2-n}{2+n}$$
Substituting $m$ in the equation $mn + 2m + 2n + 2 = n$ for $\frac{2-n}{2+n}$ and get following result: 
$$n=n$$
 A: You can factor this as:
$$mn + 2m + 2n + 2 = n \iff m(n + 2) + (n + 2) = 0 \iff (m+1)(n+2)=0$$
Which is only true if $m=-1$ or $n=-2$.
A: Hint $\ $ A polynomial $\,f\,$ on $\,\Bbb Z\,$ has more roots than its degree $\!\iff\! f = 0,\,$  i.e. all coeff's $ = 0.$ 
Hence  $\,f(n) = (m\!+\!1)\, n + 2(m\!+\!1)\,$ has $\,> 1 $ root $\iff \,m\!+\!1 = 0 = 2(m\!+1)\iff\, \ldots$
Remark $\ $ Note how this conceptual reformulation into polynomial form makes it obvious. The other proofs can be viewed as proofs of this general result for the special case of linear polynomials. 
A: Here is another approach to this kind of question.
Choose a value of $n$ which makes things simple - $n=0$ looks good as many of the terms vanish.
That gives $2m+2=0$ so that $m=-1$
Now substitute $m=-1$ in the original equation to obtain$$-n-2+2n+2=n$$This is an identity which is valid for all $n$
There can't be another solution, because it would not work for $n=0$.
Often, when you don't quite understand an equation, trying a simple value or two can help with what is going on.
A: You have solved the equation incorrectly.  You should get $$m=\frac{-n-2}{n+2}=-1.$$
As pointed out in the comments, you need to consider $n=-2$ separately.
A: First show that there is at least one $m$ with the required property, $m=-1$ does the job so that is OK.
Now suppose that $m_1\ne -1$ is such that for all integers $n$ that $m_1n+2m_1+2n+2=n$. This must be true for $n=1$, so it follows that $m_1=-1$ a contradiction. Hence $m=-1$ is the unique solution ...
A: The above equation can be re-written as:
$$(m+1)(n+2) = 0$$
The only value of $m$ that will make this true for $\forall n$ is $m = -1$.
