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!