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 $y=k+1$)
 A: 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!
A: Note that $f(f(n)) = 2n-f(n) \ge 1$ so we have $1 \le f(n) \le 2n-1$.
Hence we must have $f(1) = 1$.
It is straightforward to check that $f$ is injective.
Suppose $f(k) = k $ for $k=1,...,n$. Note that we must have $f(n+1) \ge n+1$ since $f$
is injective. Similarly, we must have $f(f(n+1)) \ge n+1$, and since
$f(f(n+1))+f(n+1) = 2(n+1)$, we must have $f(n+1) = n+1$.
A: Induction is definitely useful to show that $f(n)=n$ is the only solution. First note that $f$ has to be injective, for if $f(n)=f(m)$, the functional equation would imply $2n=2m$ and hence $n=m$. 
Now we prove by induction the statement that for all $n\leq k,\, f(n)=n$. You've done the base case. $f(1)$ has to be $1$. Now suppose that $f(n)=n$ for all $n\leq k-1$. Then by injectivity, $f(k)\geq k$. This implies $f(f(k))\geq k$ as well, because if $f(f(k))=u<k$, then we also know $f(u)=u$, so injectivity would imply that $f(k)=u$, a contradiction since $u<k$. 
Since $f(f(k))+f(k)=2k$ and both summands are $\geq k$, they must both be exactly $k$, and the induction step is completed.
A: Alternatively, let $a_k(n):=f^{\circ k}(n)$ for $k\in\mathbb{Z}_{\geq 0}$ and $n\in\mathbb{Z}_{>0}$.  Here, $f^{\circ 0}(n):=n$ and $$f^{\circ k}(n):=f\big(f^{\circ(k-1)}(n)\big)$$ for $k=1,2,3,\ldots$.  Thus, for a fixed $n\in\mathbb{Z}_{>0}$, we have
$$a_{k+2}(n)+a_{k+1}(n)-2\,a_k(n)=0$$
for every $k=0,1,2,\ldots$, with initial values $a_0(n)=n$ and $a_1(n)=f(n)$.  This is a simple recursive relation, and the solution is given by
$$a_k(n)=(-2)^k\left(\frac{n-f(n)}{3}\right)+\left(\frac{2n+f(n)}{3}\right)$$
for all $k\in\mathbb{Z}_{>0}$.  Since we have $a_k(n)\in\mathbb{Z}_{>0}$ for all $k\in\mathbb{Z}_{\geq 0}$, it must hold that the coefficient of $(-2)^k$ is $0$; that is,
$$f(n)=n\text{ for each }n\in\mathbb{Z}_{>0}\,.$$
