Number Theoritic FE( from AOPS)

[mutasimmim]

Given a function $f:N\rightarrow N$ satisfies $f(f(n))+f(n)=2n+3$ for all $n$. Find $f(n)$

[Phlembac Adib Hasan]

I'll inductively prove $f(n)=n+1$.
$\small \textbf{Base case:}$
If $f^2(1)=\min(f^2(1),f(1))\leq 2$, Then using the main equation we get $f(1)\ge 3\Longrightarrow f^3(1)\ge 7\Longrightarrow f^4(1)\leq 0$ which is impossible. So $f(1)<f^2(1)$ and $f(1)\leq 2$. Note that if $f(1)=1$, then setting $n=1$ in the main equation gives $f(1)=4$ which is absurd. Hence $f(1)=2$. Also note that $f(1)$ is the minimal value in the range of $f$.

Suppose for all $i<n$, we have $f(i)=i+1$. Since $f$ is injective, we must have $f(i)>n$ for $i\ge n$. But if $f^2(n)<f(n)$, then $f^2(n)\leq n+1\Longrightarrow f^2(n)=n+1$. So $f(n)=n+2$. But in that case $f^3(n)=n+6$ and $f^4(n)=n-1$ which is impossible from inductive hypothesis. Hence $f(n)=\min(f^2(n),f(n))\leq n+1$. So $f(n)=n+1$.

[mutasimmim]

Actually once we proved the base case, the inductive proof follows at once like this:
$f(fn))+f(n)=2n+3\Longrightarrow f(n+1)+(n+1)=2n+3$[ By induction hypothesis] . Which obviously implies $f(n+1)=n+2$.

However, here is my proof that $f(1)=2$
As we see $f(f(1))+f(1)=5$, we consider the following cases:
Case 1
$f(1)=4, f(f(1))=1\Longrightarrow f(f(1))=f(4)=1\Longrightarrow f(f(4))=10$, from the given condition but which is obviously a contradiction as $f(f(4))=f(1)=4$

Case2
$f(1)=3,f(f(1))=2\Longrightarrow f(3)=2\Longrightarrow f(f(3))=f(2)=7\Longrightarrow f(f(2))=0$, which isn't acceptable.

Case 3
$f(1)=1, f(f(1))=4$ which is of course impossible as $f(f(1))=f(1)=1$, not $4$.

So we are left with the only case $f(1)=2, f(f(1))=3$.

[mutasimmim]

$f(1)\ge 1 \Longrightarrow f^3(1)\ge 7$

Adib, why this? Didn't get it.

[Phlembac Adib Hasan]

$f^3(1)=2f(1)+3-f^2(1)\ge 2\cdot 3+3-2=7$