Is g(n)=8n/15 Only When n Is Divisible by 3 and 5?

[Poirot1]

Show that g(n)=8n/15 iff n is divisible by 3 and 5 and by no other primes, where g is the euler totient function.

How to go about the proof?

 

[Nono713]

Re: totient function proof

Rewrite $N$ as a product of powers of $3$ and $5$. Use the multiplicative property, and what is the totient of a prime power?

Full derivation below.

Spoiler

If $N = 3^m \cdot 5^n$ then $\varphi(N) = \varphi(3^m \cdot 5^n)$. Using the multiplicative property:

$$\varphi(3^m \cdot 5^n) = \varphi(3^m) \cdot \varphi(5^n)$$
And you know that for prime $p$ we have:

$$\varphi(p^k) = p^{k - 1} \cdot (p - 1)$$
And so:

$$\varphi(3^m) \cdot \varphi(5^n) = \left ( 3^{m - 1} \cdot (3 - 1) \right ) \cdot \left ( 5^{n - 1} \cdot (5 - 1) \right ) = 8 \cdot 3^{m - 1} \cdot 5^{n - 1}$$
And...

$$3^m \cdot 5^n = N ~ ~ ~ \implies ~ ~ ~ 3^m \cdot 5^n \cdot \left ( 3^{-1} \cdot 5^{-1} \right ) = 3^{m - 1} \cdot 5^{n - 1} = N \cdot 15^{-1}$$
And we conclude:

$$\varphi(N) = 8 \cdot 3^{m - 1} \cdot 5^{n - 1} = 8 \cdot N \cdot 15^{-1} = \frac{8}{15} \cdot N$$

 

[Amer]

Re: totient function proof

first n should be divisible by 3 and 5 since phi(n) is the number of the positive integers relatively prime with n
so phi(n) is integer for all n.
suppose n is divisible by another prime p, say
$$n = 3^r 5^s p^t...(1)$$
g is multiplicative
$$g(n) = g(3^r)\cdot g(5^s)\cdot g(p^t)$$

$$g(n) = 3^r \left( 1 - \frac{1}{3} \right) 5^s \left(1 - \frac{1}{5} \right)g(p^t)$$

$$g(n) = 3^r \left( \frac{2}{3} \right) 5^s \left(\frac{4}{5} \right) g(p^t)$$

$$g(n) =3^r\cdot 5^s \left( \frac{8}{15} \right) g(p^t)$$

from (1)
$$3^r\cdot 5^s = \frac{n}{p^t}$$

$$g(n) = \frac{8n}{15} \frac{g(p^t)}{p^t}$$
we want p such that $$g(p^t) = p^t$$ , that's true for 1 just

 

[chisigma]

Show that g(n)=8n/15 iff n is divisible by 3 and 5 and by no other primes, where g is the euler totient function.

How to go about the proof?

The basic formula of the Euler's Totiens Function is...

$\displaystyle \varphi (n) = n\ \prod _{p|n} (1-\frac{1}{p})$ (1)

... so that Your equation...

$\displaystyle \varphi(n)= \frac{8\ n}{15}$ (2)

... is satisfied if...

$\displaystyle \prod _{p|n} (1-\frac{1}{p}) = \frac{8}{15}$ (3)

... and that happens only for $\displaystyle n= 3 \cdot 5 = 15$ ...

Kind regards

$\chi$ $\sigma$

 

[CellCoree]

To prove that g(n)=8n/15 iff n is divisible by 3 and 5 and by no other primes, we can use the definition of the Euler totient function and the fundamental theorem of arithmetic.

First, let's define g(n) as the number of positive integers less than or equal to n that are relatively prime to n. This means that g(n) counts the number of integers that do not share any common factors with n, except for 1.

Next, let's consider the prime factorization of n. By the fundamental theorem of arithmetic, every positive integer can be uniquely expressed as a product of primes. So, we can write n as n = p1^a1 * p2^a2 * ... * pn^an, where p1, p2, ..., pn are distinct primes and a1, a2, ..., an are positive integers.

Now, let's look at the definition of g(n)=8n/15. We can rewrite this as g(n)=n * (8/15). This means that g(n) is equal to n multiplied by a fraction, where the numerator is 8 and the denominator is 15.

Using this information, we can deduce that for g(n) to equal 8n/15, n must be divisible by both 8 and 15. Since 8 and 15 are relatively prime, this means that n must be divisible by both 8 and 15, and therefore by their product, which is 120.

Furthermore, since n is divisible by 120, we can write n as n = 2^a * 3^b * 5^c, where a, b, and c are positive integers. This is because 120= 2^3 * 3^1 * 5^1, and any other prime factors of n would make it not relatively prime to 120.

Therefore, we have shown that n must be divisible by 3 and 5, and no other primes, in order for g(n) to equal 8n/15. This completes the proof.