Show that there are infinitely many integers, n for which ϕ(n) is a perfect square.

Question

I have a couple questions about trying to understand Euler's Phi function.

So I'm being asked to show there are infinitely many integers, n for which ϕ(n) is a perfect square.

I understand that the Phi function lists the number of positive integers less than or equal to n that are relatively prime to n. I also know that if p is prime then ϕ(p)=p-1

However I just have a couple questions regarding my proof:

ϕ($ 2^n$)=($ 2^n$)(1-$\dfrac{1}{2}$)=$\dfrac{2^n}{2}$=$2^{n-1}$ How do I get the final line of this proof?

and from there when I substitute odd values such as 1 I get 1, 3 I get 4, 5 I get 16, 7 I get 64 illustrating that I keep getting perfect squares.

For which $n$ is $2^{n-1}$ a perfect square, i.e., of the form $2^{2m}=(2^m)^2$? — Dietrich Burde, Dec 22 '19 at 15:59
Sorry if this question is really obvious how can we jump from n-1 to 2m? I know a perfect square needs to be raised to the power of 2 but where does the m come into place? — Lil, Dec 22 '19 at 16:01
@Lil $\ 2^k\ $ is a perfect square, if and only if $\ k\ $ is even. In general, we have a perfect square if and only if every exponent in the prime factorization is even. — Peter, Dec 22 '19 at 16:05
One law about powers is $$(a^m)^n=a^{m\cdot n}$$ which gives here $2^{n-1}$. And if $n$ is odd, then $2^{(n-1)/2}$ is an integer. — Peter, Dec 22 '19 at 16:12
We know that for prime p ,$\phi(p)=p-1$. so $p-1 $ must be perfect square. so p must be in form $n^2+1$. — sirous, Dec 22 '19 at 16:18
@sirous That is not really helpful . It is still unproven whether there exists infinite primes of the form $n^2+1$ . See this primes-of-the-form-n21-hard. — The Demonix _ Hermit, Dec 22 '19 at 16:23
See also https://math.stackexchange.com/questions/535402/are-there-infinitely-many-n-for-which-varphin-is-a-perfect-square — lhf, Dec 22 '19 at 18:10

score 3 · Answer 1 · answered Dec 22 '19 at 16:31

You are almost there. As you found, for $n$ odd, $2^{n-1}$ is a perfect square, because $n-1$ is even and $2^{n-1}$ is the square of $2^{(n-1)/2}$.

Therefore, for $n$ odd, $\phi(2^n)=2^{n-1}$ is a perfect square, and since there are infinitely many positive odd integers, we have shown that there are infinitely many integers for which the totient function is a perfect square.

score 1 · Answer 2 · answered Dec 22 '19 at 16:15

1

Another example is $n=5^{2m+1}$. Then $\phi(n)=5^{2m}\cdot 4=(5^m\cdot 2)^2$.

This argument works for all primes of the form $2^{2k}+1$, that is, all Fermat primes: $5, 17, 257, 65537$ are the only ones known to date.

answered Dec 22 '19 at 16:15

lhf

216,483

3

Indeed, for all primes of the form $4n^2+1$, such as $p=37$ or $p=101$. – Angina Seng Dec 22 '19 at 17:01
@LordSharktheUnknown, but are there infinitely many of those? https://oeis.org/A121326 – lhf Dec 22 '19 at 17:02
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@LordSharktheUnknown It seems that it is incredibly hard to show that there are infinite prime of form $x^2+1$ . primes-of-the-form-n21-hard. – The Demonix _ Hermit Dec 22 '19 at 17:09

Show that there are infinitely many integers, n for which ϕ(n) is a perfect square.

2 Answers2