4

(I want to clear out the fact that I'm no professional mathematician. Please feel free to correct me if needed. I have no mathematical background. I am 13 years old, this question is originated from pure math curiosity)

Let $\mu$ be any number greater than zero, such that $N_d^{\mu}$ is a natural number.

Let $N_d$ be any natural number with $d$ divisors. $N_d$ has to be a prime number (or a power of prime).

Is it correct to say that $N_d^{\mu}=N_{\left(d-1\right)\mu+1}$ ?

If so, why?

An small example: let $N_d = 7$. That means $d = 2$, since the number $7$ has two divisors. We can know that the result of the expression $(d-1)\mu+1$ results in the amount of divisors that $7^\mu$ has.

Let $\mu=5$ for this example.

We can know that $7^5$ (which is equal to $16807$) has $(2 - 1)5+1=6$ divisors.

And that is correct: $16807 = 1*7*49*343*2401*16807$

guilherme1771
  • 43

1 Answers1

4

If a number $k$ has a prime factorisation $p_1^{n_1}.p_2^{n_2}.p_3^{n_3}..... $ then it has $d=(n_1+1)(n_2+1)(n_3+1)..... $ divisors.

Similarly $k^{\mu} $ have $d'=(\mu n_1+1)(\mu n_2+1)(\mu n_3+1)..... $ divisors.
You can see that in general that $d'\ne(d-1)\mu+1$ however equality holds for all $k $ which is a prime or a power of a prime.

You can infact prove that equality holds only if k is a prime or a power of a prime.

Asher2211
  • 3,406
  • 10
  • 31