Let $P(x,y)$ be the assertion $f(x)+f(y) \mid (x+y)^{k}$.

**(1)** $f(1)=1$

**Proof:**First assume that $f(1) \neq 1$.$P(1,1)$ implies $f(1)=2^{a}$ for some non-negative integer $a$.

$P(2,2)$ implies $f(2)=2^{b}$ for some non-negative integer $b$.

$P(1,2)$ implies that $f(1)+f(2) \mid 3^{k}$.If both $f(a),f(b)$ are even,then we get a contradiction as an even integer doesn't divide an odd integer.So,one of them is odd i.e. equals to $1$.As we assumed that $f(1) \neq 1$,so,$f(2)=1$.

So,$2^{a}+1=3^{r}$........**(i)**,then by **Zsigmondy's Theorem**,$a=1$ or $3$.

Now,$P(1,3)$ implies that $f(1)+f(3) \mid 4^{k}$.So,$f(1)+f(3)=2^{i}$ where $i$ is a positive integer.Since $f(1)$ is a power of $2$,so,$f(1)=f(3)$.

Now,$P(2,3)$ implies that $1+2^{a} \mid 5^{k}$ $\Rightarrow$ $2^{a}+1=5^{r}$.......**(ii)**.From **(i)**,we know that $a=1$ or $3$,but none is valid for **(ii)**,a contradiction.So our first assumption was wrong and $f(1)=1$.

**(2)** $f(p-1)=p-1$

**Proof:**$P(1,p-1)$ implies that $f(p-1)=p^{s}-1$.

$P(p-1,p-1)$ implies that $f(p-1)=p^{s}-1 \mid 2^{k-1}(p-1)^{k}$.So ,the set of distinct prime divisors of $p-1$ and $p^{s}-1$ is same.But from **Zsigmondy's Theorem**,we know that this can be possible if and only if $s=1$ i.e. $f(p-1)=p-1$.

**(3)** $f(n)=n$ for all $n \in \mathbb{N}$

**Proof:**$P(n,p-1)$ implies that $f(n)+p-1 \mid (n+p-1)^{k}$.

Now,$f(n) \equiv -(p-1)$ $\text{(mod}$ $f(n)+p-1)$.

$\Rightarrow$ $n-f(n) \equiv n+p-1$ $\text{(mod}$ $f(n)+p-1)$.

$\Rightarrow$ $(n-f(n))^{k} \equiv (n+p-1)^{k} \equiv 0$ $\text{(mod}$ $f(n)+p-1)$

Here the value of $(n-f(n))^{k}$ is fixed for each particular $n$.But if $p \to \infty $, $(f(n)+p-1) \to \infty$.Therefore,$(n-f(n))^{k}=0$ and $f(n)=n$ for all $n \in \mathbb{N}$.