Are there infinitely many $n$ for which $\varphi(n)$ is a perfect square?

Question

Prove or disprove: $\phi(n)$ is a perfect square for only a finite number of odd numbers n.

I know it works for even numbers since we can use $n=p^k$ and have $p=2$, however, I don't know about odd numbers.

Answer 1

Consider $5^{2m+1}17^{2n+1}$. Similarly, one could use $3^{2m+1}19^{2n+1}$. One can play the same game with $11$ and $41$. Or use $3^{2a+1}7^{2b+1}11^{2c+1}31^{2d+1}$.

Remark: It has been conjectured that there are infinitely many primes of the shape $x^2+1$. There is considerable numerical evidence that the conjecture is true. In that case there are infinitely many primes $p$ such that $\varphi(p)$ is a perfect square.

Answer 2

The process of finding perfect squares among odd numbers, is similar to that of finding things perfect numbers. I should imagine there are an infinite number of these things, though.

For example, the totient of 7 is 6. You then just have to find eg a prime of the form $6n^2+1$, such as $97$, and the totient of $7\cdot 97$ is itself a square. Also, since $601 = 6\cdot 10^2+1$, so is the product of any two of these numbers, like $4207$, which gives $60^2$. These can be multiplied by any even power, such as $7^3$ has a totient of $7^2 \cdot 6$.

One notes that any prime of the form of $n^2+1$ has a square totient, as does multiplying this by any of the numbers found above.

Even if one supposes powers, one can find examples, like $3^2$, whose totient is 6. Since these are identical with the previous example, then the totient of say $63$ is $6^2$, and so forth.

Answer 3

Take any $p=n^2+1$, $p$ is prime, $n>2$ is natural number. It's clear that $p$ is odd. Now, I have a lemma stating that $ \forall p = n^2+1, \phi(p^{2k+1})=(p-1)*p^{2k}$. Let's prove this lemma by induction:

For $k=0$ and $p=n^2+1$, as $p$ is prime, $\phi(p^{2k+1}) = \phi(p)= p-1$

Now, assume the lemma is true with k. Check out what's gonna happen with k+1?

$p^{2(k+1)+1} = p^{2k+3}$

Think about the numbers which divides $p^{2k+3}$. As $p$ is prime, the only numbers which are relatively prime with $p^{2k+3}$ are the ones that can be divided by $p$. We know that the number of the ones which are smaller than $p^{2k+1}$ is $(p-1)*p^{2k}$ with the assumption of induction. And the number of numbers between $p^{2k+1}$ and $p^{2k+3}$ is $p^{2k+3}-p^{2k+1}$, in total. The ratio of being relatively prime with $p^{2k+3}$ is $(p-1)/p$. Then the number of numbers which are not relatively prime with $p^{2k+3}$ and between $p^{2k+1}$ and $p^{2k+3}$ is:

$$[(p^{2k+3}-p^{2k+1})*(p-1)]/p$$ $$= [p^{2k+1}*(p^2-1)*(p-1)]/p $$ $$= p^{2k} *(p-1)^2 * (p+1) $$

Now add this to the number of the ones that smaller than $p^{2k+1}$, which is $(p-1)*p^{2k}$, we have:

$$[(p-1)*p^{2k}] + p^{2k} *(p-1)^2 * (p+1)$$ $$ =p^{2k}*(p-1)*[(p-1)*(p+1)+1] $$ $$ = p^2k*(p-1)*[p^2-1+1]$$ $$ = p^{2k}*(p-1)*p^2$$ $$ = p^{2k+2}*(p-1)$$ $$ = p^{2(k+1)}*(p-1)$$

This is the proof of the lemma.

Now, it's clear that $(p-1)*(p^{2k})=n^2*p^{2k}$ is a perfect square. If we assume that some number m is the maximum number such that $\phi(m)$ is a perfect square, because we could find a number $p^{2k+1} > m$ in every time, this assumption would be wrong. Then, ϕ(n) is a perfect square for infinitely many numbers of odd numbers n.

Example: $5$ is a prime number and $5=2^2+1$. Then, $\phi(5)$, $\phi(5^3)$, $\phi(5^5)$,.....,$\phi(5^{2k+1})$, .... are all perfect squares and they are infinitely many.