Conjecture:Today I have no intention of thinking about this question. I have only got two solutions so far. I guess there are only two solutions, but I won't prove it.

Let $n$ be positive integers, such that $$n+\tau{(n)}=2\varphi{(n)}$$ where $\varphi$ is the Euler's totient function and $\tau$ is the divisor function i.e. number of divisors of an integer.

It is clear $n=1$ works,and also I found out $n=9$ is another answer, because $\tau{(9)}=3$, $\varphi(9)=9\left(1-\dfrac{1}{3}\right)=6$, so we have $$9+3=2\cdot 6\Longleftrightarrow 9+\tau{(9)}=2\varphi(9)$$

But how to find others? I tried a lot, but I couldn't find any more.

  • $\begingroup$ It should be clear that n is an odd square. What have you tried? Gerhard "Is Sometimes An Odd Square" Paseman, 2018.03.20. $\endgroup$ Mar 20, 2018 at 14:55
  • 2
    $\begingroup$ Croossposted to MSE. $\endgroup$ Mar 20, 2018 at 20:02

1 Answer 1


First of all, if $n$ is even, then $\varphi(n)\leq n/2$, so $n\geq 2\varphi(n)$. Therefore, $n$ is always odd and so is $\tau(n)$. Thus, as Gerhard noticed, $n=m^2$ for some odd integer $m$. Consequently, $m^2+\tau(m^2)=2\varphi(m^2)=2m\varphi(m)$. Thus, $\tau(m^2)$ is divisible by $m$. Now, let us notice that

$$\tau(m^2)=2\sum_{\substack{d \mid m^2 \\ d<m}} 1+1\leq 2m-1.$$

Therefore, we have


Let $m=p_1^{\alpha_1}\ldots p_\ell^{\alpha_\ell}$ for some odd primes $p_i$ and $\alpha_i\geq 1$. Now, for any $p$ and any $\alpha\geq 1$ we have

$$p^{\alpha}\geq 1+\alpha(p-1),$$

thus for any odd prime $p$ and any integer $\alpha>0$ we have

$$p^{\alpha}\geq 1+2\alpha$$

and equality is attained only if $p=3$ and $\alpha=1$. Therefore, we get either $m=3$ or $m=1$, which gives exactly the solutions you already mentioned.

  • 6
    $\begingroup$ There is a more elegant way to finish. In your first display, each $d$ is odd, hence in fact we get $\tau(m^2)\leq m$. Therefore, we have equality in this bound, so each odd $d<m$ divides $m^2$. For $m>1$, this implies that $m-2\mid m^2$, hence also $m-2\mid 4$. As $m$ is odd, $m-2=1$, whence $m=3$. $\endgroup$
    – GH from MO
    Mar 21, 2018 at 3:12

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