If $$n=p_1^{a_1}\cdots p_k^{a_k},$$ then define

$$f(n):=p_1^2+\cdots+p_k^2$$

So, $f(n)$ is the sum of the squares of the prime divisors of $n$.

For which natural numbers $n\ge 2$ do we have $f(n)=n$ ?

It is clear that $f(n)=n$ is true for the square of any prime, but false for the other prime powers.

If $p$ and $q$ are the only prime divisors of $n$, we would get $p^2+q^2\equiv 0\pmod p$, which implies $p=q$, so for numbers with exact two prime divisors, $f(n)=n$ cannot hold.

If $p,q,r$ are primes with $p<q<r$, then we have two possibilities.

If $p,q,r\ne 3$, we have $p^2+q^2+r^2\equiv 0\pmod3$, so $f(n)=n$ cannot hold. If $p=3$ or $q=3$, then $p^2+q^2+r^2 \equiv 2\pmod3$, so $p^2+q^2+r^2$ is not divisible by $3$, so $f(n)=n$ cannot hold.

Finally, if $p<q<r<s$, then if $p>2$, then $p^2+q^2+r^2+s^2\equiv 0\pmod4$, so $f(n)=n$ cannot hold. And if $p=2$, then $p^2+q^2+r^2+s^2\equiv 3\pmod4$, so $p^2+q^2+r^2+s^2$ is odd and $f(n)=n$ again cannot hold.

So, apart from the squares of the primes, the number must have at least $5$ prime factors. I searched to about $6\times 10^7$ and did not find a "non-trivial" example.

- Is there a number $n$ with at least two prime factors and $f(n)=n$ ?