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.