Proof verification: system of functional equations: $f(0)=1$, $f\bigl(f(n)\bigr)=n$ and $f\bigl(f(n+2)+2\bigr)=n$ A problem in Putnam Competition 1992(?). The question asked:

Prove that, the only solution of the system of functional equation with respect to $f:\mathbb Z\to\mathbb Z$:$$
\begin{cases}
f\bigl(f(n)\bigr)=n\\
f\bigl(f(n+2)+2\bigr)=n\\
f(0)=1
\end{cases}
$$
is $f(n)=1-n$.

I know that the usual way is to construct another solution, and show that the difference between the two solutions is zero. But I want to use equivalence condition this time. (i.e. Prove that the system is equivalent to say that $f(n)=1-n$).
Now, we have
$$f\bigl(f(n)\bigr)=f\bigl(f(n+2)+2\bigr)$$
Take $f$ on both sides.
$$f\Bigl(f\bigl(f(n)\bigr)\Bigr)=f\Bigl(f\bigl(f(n+2)+2\bigr)\Bigr)$$
Again by the first equation,
$$f(n)=f(n+2)+2$$
And another useful thing is that,
$$f(1)=f\bigl(f(0)\bigr)=0$$
So, the original system is equivalent to say the recurrent relation
$$
\begin{cases}
f(n)=f(n+2)+2\\
f(0)=1\\
f(1)=0
\end{cases}
$$
The sequence-like function  satisfying above is unique, trivially (as for all integers, we could find an unique image of it). So our proof is actually done already as $f(n)=1-n$ is simply a solution. Q.E.D.
In my proof, I used a lot of equivalence condition. I know that my proof is valid if the conditions are actually equivalent. I think so for myself, but I want peer reviews also. Thanks in advance.
 A: Your proof is correct, well done! If you wanted to, you could show that from your last recurrence relation $$
\begin{cases}
f(n)=f(n+2)+2\\
f(0)=1\\
f(1)=0
\end{cases}
$$ yields a unique solution through a simple induction as well. 
A: Note that $f$ is an injective function.  To show this, suppose $f(m)=f(n)$ for some $m,n\in\mathbb{Z}$.  Therefore,
$$m=f\big(f(m)\big)=f\big(f(n)\big)=n\,.$$
Now, the condition $f\big(f(n+2)+2\big)=n$ comes into play.  Observe that
$$f\big(f(n)\big)=n=f\big(f(n+2)+2\big)\,.$$
By injectivity,
$$f(n)=f(n+2)+2\text{ for all }n\in\mathbb{Z}\,.\tag{*}$$
Since $f(0)=1$, we conclude by induction on $n$ that
$$f(2n)=1-2n\text{ for every }n\in\mathbb{Z}_{\geq 0}\,.$$
We can also induct the other way around, using (*), again to show that
$$f(2n)=1-2n\text{ for every }n\in\mathbb{Z}_{<0}\,.$$
This shows that $f(n)=1-n$ for all even integers $n$.  Now, let $n$ be an odd integer.  We get that
$$f\big(f(n)\big)=n=1-(1-n)=f(1-n)\,,$$
as $1-n$ is an even integer.  Therefore,
$$f(n)=1-n\text{ for each odd integer }n$$
as well.
