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This question was inspired by this MSE question.

In MSE, it is shown that $$n - \varphi(n) = (2^{p-1})^2$$ if $n = {2^{p-1}}(2^p - 1)$ is an even perfect number.

Here is my question in this post:

Is $N - \varphi(N)$ a square, if $N = q^k m^2$ is an odd perfect number with special prime $q$?

MY ATTEMPT

Note that $$N - \varphi(N) = q^k m^2 - \varphi(q^k) \varphi(m^2) = q^k m^2 - (q^k - q^{k-1})(m\varphi(m)) = q^{k-1} m (qm - (q - 1)\varphi(m)).$$ Since the Eulerian form $N = q^k m^2$ of an odd perfect number dictates that $q \equiv k \equiv 1 \pmod 4$, then we test whether $$m (qm - (q - 1)\varphi(m))$$ is a square. Suppose otherwise.

Then $$m (qm - (q - 1)\varphi(m)) = Q^2$$ which forces $q \equiv 1 \pmod 8$, since $\varphi(m)$ is always even. (Since $q$ is the special prime, this means that $q \geq 17$.)

Alas, this is where I get stuck.

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    $\begingroup$ This is likely a very tough question. Notice that if one could prove that $N-\phi(N)$ is a perfect square this would imply that $N$ is not 5 (mod 8), since $8|\phi(n)$. However, this would among other things, imply that 5 is not the special prime. Thus, it is very likely that proving this is well outside what current techniques can do. $\endgroup$
    – JoshuaZ
    Commented May 27 at 20:04
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    $\begingroup$ Note also that the statement is at least false if the largest prime divisor of $N$ is not $q$. If the largest prime divisor is $p \neq q$, then $p^a||n$ for some even $n$. But $p^{a-1}|| n - \phi(n)$ and so $n$ is not a perfect square. So this would also force that the largest prime divisor of $n$ must be the special prime. Overall, it seems like a more productive direction to go in would be to try and show that $n - \phi(n)$ is NOT a perfect square, but I don't see any obvious way to do that either. $\endgroup$
    – JoshuaZ
    Commented May 27 at 20:08
  • $\begingroup$ @JoshuaZ: I appreciate your comment, Josh! I was hoping someone with more experience on the Euler totient function would be able to settle this one. (This question was mainly triggered by an inquiry that I received via e-mail.) $\endgroup$
    – Jose Arnaldo Bebita
    Commented May 27 at 20:08

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