Strange Function

The result $d(p_{1}^{a_{1}}....p_{k}^{a_{k}})=(a_{1}+1)(a_{2}+1).....(a_{k}+1)$ will be used throughout

Putting $x=1$ in (i), $d(f(1))=1$ . Hence, $f(1)=1$ .

Note that $f$ is one-one

Now, $d(f(p))=p$ for prime p. Hence, $f(p)=q^{p-1}$ for some prime q. Hence, $f(2)=q^{2-1}=q$ , a prime.

Suppose, $p>2$ be a prime.

Putting $x=2$ and $y=p$ in (ii) we get

$f(2p)|p^{2p-1}f(2)$

Putting $x=p$ and $y=2$ in (ii) we get

$f(2p)|(p-1)2^{2p-1}f(p)=(p-1)2^{2p-1}q^{p-1}$

Suppose $p≠q$ .Hence $p$ does not divide $(p-1)2^{2p-1}q^{p-1}$ .Hence the greatest common factor of $p^{2p-1}f(2)$ and $(p-1)2^{2p-1}q^{p-1}$ is a divisor of $f(2)$ .Since $f(2p)$ is a common divisor of them, $f(2p)$ divides $f(2)$ . But $f(2)$ is a prime and $f(2p)>1$ . So, $f(2p)=f(2)$ . contradiction!

So, $p=q$ .Therefore, $f(p)=p^{p-1}$

Suppose p=2. Putting $x=2$ and $y=3$ in (ii) and again, $x=3$ and $y=2$ in i(i), we get $f(6)|3^{5}f(2)$ and $f(6)|2^{6}f(3)$ .Suppose $f(2)$ is odd. Then, $f(6)|f(3)=9$ .Therefore, $f(6)={1,3,9}$ .Then, $6=d(f(6))={d(1),d(3),d(9)}={1,2,3}$ , which is false. So, $f(2)=2$

Next, for each $n>1$ , the prime divisors of $f(n)$ are among the ones of $n$ .Indeed, let $p$ be the least prime divisor of $n$ .Putting $x=p$ and $y=\frac{n}{p}$ in (ii), we get $f(n)|(p-1)y^{n-}p^{p-1}$ .Write $f(n)=lP$ where $gcd(l,n)=1$ and $P$ is a product of primes dividing $n$ .Since, $l$ divides $(p-1)y^{n-}p^{p-1}$ and is co-prime to $y^{n-1}p^{p-1}$ , $l$ divides $p-1$ .So, $d(l)≤l<p$ . Now, $n=d(f(n))=d(lP)=d(l)d(P)$ .Hence, $d(l)|n$ .But $d(l)<p$ ,the least prime divisor of $n$ .So, $l=1$ , proving the claim.

If $p$ is a prime and $a≥1$ ,by the above, the only prime factor of $f(p^{a})$ is $p$ .So, $f(p^{a})=p^{b}$ .Hence, $p^{a}=d(f(p^{a})=d(p^{b})=b+1$ .Therefore, $f(p^{a})=p^{b}=p^{p^{a}-1}$ .

Finally, show that if $n=p_{1}^{a_{1}}....p_{k}^{a_{k}}$ , $f(n)=p_{1}^{p_{1}^{a_{1}}-1}....p_{k}^{p_{k}^{a_{k}}-1}$ .

Now, calculate the given sum.