Question

There are infinitely many n with phi (n) odd. True or False
Please, help

Answer 1

define phi(n)

Answer 2

relative prime function ... the number of integers less than or equal to 'n' that are relatively prime to n

Answer 3

I don't get it. Please, explain

Answer 4

you presented the totient function i believe, so i asked you to define it

Answer 5

surely if we have p is a prime, then phi (p) = p-1 is an even number. But that doesn't mean the number of odd is finite.

Answer 6

i then looked it up to make sure i was on the right track

Answer 7

i see a property of the totient function from the wolf but im not sure how to prove it.

Answer 8

oh, I think I got it

Answer 9

if n> 2, then phi (n) is even

Answer 10

thats the property.

Answer 11

hence we have only 1case gives us phi (n) odd, that is phi (2) =1. Therefore, the statement is false. Am I right?

Answer 12

phi(0) = 1
phi(1) = 1
phi(2) = 1

regardless tho, the rest are even so there is only a finite number of odd phis

Answer 13

I don't get phi(0) and phi (1) . why?

Answer 14

I'm seeing this (see attached)

from this link

https://en.wikipedia.org/wiki/Euler%27s_totient_function

Answer 15

wolf says its conventional

Answer 16

@jim_thompson5910 Thanks for the link.
@amistre64 I can't say "wolf says its conventional" on my test. I need logic.

Answer 17

so yeah that's the same as what you said when you wrote `if n> 2, then phi (n) is even `

Answer 18

which is why i asked you to define the function to start with.

Answer 19

In my class, phi (n) = number of relative prime with n from residue set

Answer 20

But we count from 1 to n -1, not 0 there

Answer 21

your material should have phi(0) and phi(1) defined as well.

oh, then they should at least have defined phi(1) in the material

Answer 22

Like if n = 5, then the residue set is {1,2,3,4} and phi (5) = 4.

Answer 23

it matches to phi (prime) = prime -1

Answer 24

from that article I posted, it says

phi(m*n) = phi(m) * phi(n)

and

phi(p^k) = p^k*(p-1)

so effectively, for any integer n, you can factor it into its prime factorization

You'll get a bunch of things that look similar to p^k*(p-1) being multiplied out
since p is prime, p is odd and p-1 is even

so p-1 is a factor of n, making the product of terms even overall

Answer 25

sorry I meant to write p^(k-1)*(p-1)

Answer 26

Yes, I got you. Thank you so much.
I am looking for phi (0) and phi (1) as what asmitre said. I didn't see them yet

Answer 27

and I also meant to say p-1 is a factor of phi(n)

Answer 28

this "phi(m*n) = phi(m) * phi(n)"

is not true


you need (m,n)=1

Answer 29

right, I missed that part, my bad

Answer 30

Yes, I saw it also. :) (from my book)

Answer 31

Thank you @Zarkon