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 postive integers, such that $$n+\tau{(n)}=2\varphi{(n)}$$ where φ is the Euler's totient function and τ 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.

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.

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

it is clear $n=1$,and also I found $n=9$ 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 other? I tried a lot. I couldn't find any more.

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 postive integers, such that $$n+\tau{(n)}=2\varphi{(n)}$$ where φ is the Euler's totient function and τ 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.