Find all functions $f:\mathbb{N}^+\to\mathbb{N}^+$ such that $f\big(f(n)\big)+f(n)=2n$ for every $n\in\mathbb{N}^+$. 
Find all functions $f:\mathbb{N}^+\to\mathbb{N}^+$ such that $$f\big(f(n)\big)+f(n)=2n$$ for every $n\in\mathbb{N}^+$.

I think the answer is $f(n)=n$,We prove this by induction. (at last step I can't induction it）

It is true for  $n=1$ because
Let $f(1)=x\ge 1,$and let $n=1$,we have
$$f(x)+x=2\Longrightarrow x=1$$

(2) Suppose it is true for
$f(n)=n(n\le k)$,
then for $n=k+1$,let $f(n+1)=y$,so we have
$$f(y)+y=2(k+1)\Longrightarrow f(y)=2(k+1)-y$$
it is easy to have $k+1\le y\le 2k+1$,and
$$f(f(y))+f(y)=2y\Longrightarrow f(2k+2-y)=2y-f(y)=3y-2(k+1)$$
since $2k+2-y\in [1,k+1]$,then I can't it ,can you help? Thanks
(if $2k+2-y\in [1,k]$,I have done it! bacuase $f(2k+2-y)=2k+2-y$,so $y=k+1$)

At first, we will show that $f$ is injective.
Suppose, for some $m,n$
$f(m)=f(n)\implies f(f(m))=f(f(n))$ therefore, by the given condition we have, $m=n$
Now, we claim, $f(n+1)\ge n+1$
Otherwise, 
suppose $f(n+1)=n+1-k (1\ge k\le n)=f(n+1-k)$ (by induction)
Therefore, $n+1=n+1-k$ (as $f$ is injective)
So, a contradiction arise.
Hence, $f(n+1)\ge n+1$
In a similar way we can show that $f(f(n+1)\ge n+1$
Hence, $2(n+1)=f(n+1)+f(f(n+1))\ge 2(n+1)$
Therefore, we must have, $f(n+1)=n+1$ and hence we are done!
