show that there are infinitely many integers $n$ such that $φ(n)\equiv2\pmod4$, where $φ(n)$ is Euler's totient function

Question

Let $p$ be a prime and $k$ be an integer greater or equal to $1$. Then $φ(p^k) = p^k - p^{k-1}$.

Answer 1

Hint:

For any prime $p\equiv 3\mod 4$, you have $\varphi(p)=p-1\equiv2\mod4$.

Answer 2

Hint: $\phi(n)$ is even for $n > 2$.

Answer 3

For one thing, $\varphi(p)\equiv 2\pmod 4$ for all primes $p$ such that $p\equiv 3\pmod 4$. The set of such primes is infinite.

Answer 4

Excepting $2$ and $3$, primes are of the form $p=6k\pm1$. Since $\varphi(p)=p-1$ for a prime, we have $\varphi(p)=6k$ or $\varphi(p)=6k-2$ for such primes.

For $p=6k+1$, we require $\varphi(p)=6k\equiv2\pmod{4}$ which is satisfied when $k$ is odd. As there are an infinite number of primes of the form $12j+7$ (equivalently $12j-5$) we are done.

For $p=6k-1$, we require $\varphi(p)=6k-2\equiv2\pmod{4}$ or equivalently $6k\equiv4\equiv0\pmod{4}$, which is satisfied when $k$ is even. As there are an infinite number of primes of the form $12j-1$ we are done.

To complete the proof you must show, as stated, that there are an infinite number of primes of the form $12j+7$ and $12j-1$, note this boils down to proving an infinitude of primes $p$ satisfying $p\equiv3\pmod{4}$.