Final answer:
There are infinitely many n for which ϕ(n) is a perfect square by considering powers of 4. Taking n as 4^k for any non-negative integer k, ϕ(n) gives a perfect square, 2^{2k-1}, since k can be infinitely large.
Step-by-step explanation:
The problem at hand is to show there are infinitely many n for which the Euler's totient function ϕ(n) is a perfect square. A strategy to demonstrate this is by considering n to be a power of 4, specifically n = 4^k for some non-negative integer k. Since ϕ(n) for a prime power like p^k (where p is a prime number) is p^k-p^{k-1}, and 4 is 2 squared, ϕ(4^k) equals 2^{2k} - 2^{2k-1}, which simplifies to 2^{2k-1}(2-1) = 2^{2k-1} which is indeed a perfect square since 2k-1 is an integer. Therefore, there are infinitely many such n, because there are infinitely many non-negative integers k.