Functional equation (N to N)

Question

Find all $f : \mathbb{N} \to \mathbb{N}$ which satisfy the equation:

$f(d_1)f(d_2)...f(d_n)=N$

Where $N$ is a natural number and $d_i, 1 \leq i \leq n$ are all of the divisors of $N$.

From what you have given, $1 = f(1)$. Further, $f(p)=p$ for every prime p, since $p = f(1)f(p)$. Then, product of distinct primes, $p_1p_2 \ldots p_n$ will all have function value $1$. See if this helps. — Sarvesh Ravichandran Iyer, Oct 22 '16 at 09:48
Thank you, I have found that f(2^n)=2. As a natural number can be written as a multiplication of some power of 2 and some power of certain primes, then he function should give each prime^n value of that prime and each power of 2 value of 2, all other numbers should get value of 1? — J.Does, Oct 22 '16 at 10:11

score 1 · Answer 1 · answered Oct 22 '16 at 10:29

Here is my attempt.

Any $N$ will have a prime power factorisation, so its divisors are all powers of primes, or a product of prime powers.

Consider $N$ = $p^2$ with $p$ prime. Taking $f(p) = p$ and $f(1) = 1$ as given, we have $f(1)f(p)f(p^2)=p^2$, and so $f(p^2) = p$. Generalising, $f(p^k)=p$, for any positive integer $k$.

Now when we have more than one prime in the prime power factorisation of $N$, and at least one of those primes has multiplicity greater than $1$, there will be divisors of $N$ which are not prime powers (composite numbers with more than one prime factor). e.g. $pq$ divides $p^2q^2$. But $f$ acting on these divisors must give $1$, since otherwise there is no way that $f(c_1)f(c_2)$ can equal $1$, where $c_1$ and $c_2$ are composite numbers, each with more than one prime factor. To see this, recall that the $f$ outputs natural numbers, and $1 \times 1 = 1$ is the only way to obtain $1$ from the product of two natural numbers.

Summarising: $f(1) = 1$, $f(p^k) = p$, for $p$ prime and any positive integer $k$. For all natural numbers $l$ with more than one prime factor, $f(l) = 1$.

Is it a legtitimate solution? Should I find the exact form of these functions? — J.Does, Oct 22 '16 at 10:41

score 1 · Accepted Answer · answered Oct 22 '16 at 10:32

Define $f$ as follows: Let $$f(n) = \begin{cases} p & n=p^k \text{ for some prime } p \\ 1 & \text{otherwise (including $n = 1$)}\end{cases}$$.

Then, let $N = \displaystyle\prod_{i=1}^n p_i^{r_i}$. Then, note that while taking the product of $f$ of divisors of $N$, all terms have more than one prime will vanish. So, only left behind will be the following product: $$ f(p_1) \ldots f(p_1^{r_1}) f(p_2) \ldots f(p_2^{r_2}) \ldots f(p_n) \ldots f(p_n^{r_n}) $$ But then, the value of each term is the prime inside, so we actually get this equal to $N$.

This also shows that the function is unique.

Functional equation (N to N)

2 Answers2