# Euler's Totient function

• Apr 18th 2012, 01:22 AM
mi986
Euler's Totient function
The problem is to show that there are infinitely many odd integers k for which φ(k) is some number squared.

Now, I know that there are an infinite amount of positive integers that would make φ(k) a squared number .
Taking k=22m+1, where m is a non-negative integer:
φ(22m+1)=22m+1-22m=22m(2-1)=(2m)2

But I'm having trouble finding how to show that this is true for odd k.

I wrote out a list for φ of odd numbers, and found that this works for k = 5, 17, 37, 57, 63, 85, 185, 629, 2255, but I don't see a pattern emerging.

I know that IF there were an infinite amount of primes of the form n2+1, then I could take k=n2+1 and the result would immediately follow since φ(p)=p-1, where p is a prime. But I'm almost certain I'm not allowed to use that conjecture. :P

Does anyone have some suggestions as to what else I should try?

• Apr 18th 2012, 08:04 AM
Petek
Re: Euler's Totient function
Consider k = 5m17n for suitable choices of m and n.

ETA: On second thought, just consider k = 5m for suitable m.
• Apr 18th 2012, 08:32 PM
mi986
Re: Euler's Totient function
Oooh!
So, it's just like the case with 2!
take k=52m+1, where m is a non-negative int. Then,
φ(52m+1)
=52m+1-52m
=52m(5-1)
=(5m2)2

Ah, this means I can get similar results for 17, 37, 101, etc...
That's actually kind of awesome.

Thanks so much. (Happy)