Putting $n=1$, we get $f(f(1))+f(2)=3$. Thus there are two possibilities. Either $f(f(1))=1, f(2)=2$ or $f(f(1))=2, f(2)=1$. Also, we observe that $f(f(n))=n+2-f(n+1)\leq n+1$ and $f(n+1)=n+2-f(f(n))\leq n+1$. How do we get the values of $f(1)$ and $f(2)$?
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yes that's one of the possibilities – Debashish Jun 02 '19 at 07:23
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Oh thanks.. I have edited now – Debashish Jun 02 '19 at 07:25
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5This functional equation has been investigated here: Functions satisfying $f:\mathbb{N}\rightarrow\ \mathbb{N}$ and $f(f(n))+f(n+1)=n+2$ – Martin R Jun 02 '19 at 07:33
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2It was demonstrated in that question that $f(2)=1$ is not possible. – Martin R Jun 02 '19 at 07:41
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@MartinR thanks for the link. I am getting closer to the answer – Debashish Jun 02 '19 at 08:40