Consider k = 5^{m}17^{n} for suitable choices of m and n.
ETA: On second thought, just consider k = 5^{m} for suitable m.
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=2^{2m+1}, where m is a non-negative integer:
φ(2^{2m+1})=2^{2m+1}-2^{2m}=2^{2m}(2-1)=(2^{m})^{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 n^{2}+1, then I could take k=n^{2}+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?
Thanks for your time!
Oooh!
So, it's just like the case with 2!
take k=5^{2m+1}, where m is a non-negative int. Then,
φ(5^{2m+1})
=5^{2m+1}-5^{2m}
=5^{2m}(5-1)
=(5^{m}2)^{2}
Ah, this means I can get similar results for 17, 37, 101, etc...
That's actually kind of awesome.
Thanks so much.