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.

  • $\begingroup$ Upto $n=10^8$, the only solutions are $1$ and $9$. $\endgroup$ – Peter Mar 20 '18 at 13:26
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    $\begingroup$ Crossposted to MO. $\endgroup$ – Dietrich Burde Mar 20 '18 at 20:02
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    $\begingroup$ I'm voting to close this question as off-topic because it is both unclearly written and was crossposted to MO on the same day it was posted here. $\endgroup$ – Carl Mummert Apr 1 '18 at 15:03

$n=1,9$ are the only solutions.

For $n$ even, the LHS is strictly larger than the RHS.

If $n$ is odd, then we get that $\tau(n)$ is odd, so $n$ is a perfect square.

Say $n= \prod p_i^{a_i}$, in particular all $a_i\geq2$. We have $\prod p_i^{a_i-1} \mid \tau(n)$ so $$n \geq 3^{\sum (a_i-1)(p_i-1)}$$ (In general, if $\tau(n)$ is a product of primes then $n$ is at least $q$ to the power the sum of those primes minus $1$, where $q$ is the smallest prime dividing $n$.)(*)

Yet $3^{(a-1)(p-1)}\geq(p^2)^{a-1}\geq p^a$ for prime $p\geq 3$ and $a\geq 2$.

Equality can only occur if $3$ is the only prime dividing $n$, with exponent $a$ satifsying $2(a-1)=a$, i.e. $a=2$, $n=3^2$.

Note how in the odd case the approach shows that $ \varphi(n)$ does not divide $n\pm2^k\tau(n)$ except possibly for $n=1,9$.

(*) Let $\tau(n) = \prod (a_i+1)= \prod r_j$ with the $r_i$ prime.

  1. Show by induction that $\left( \prod b_i \right)-1 \geq \sum( b_i-1)$ for $b_i \geq 1$. (Or geometrically: the LHS counts lattice points in a hyperrectangle and excludes the origin; the RHS counts only the points on the coordinate axes, minus the origin.)
  2. Use $$n = \prod p_i^{a_i} \geq q^{\sum ((a_i+1) - 1)}$$ and factor each $a_i+1$ into prime factors $r_j$, using the above inequality as a lower bound, to get: $$\geq q^{\sum (r_j-1)}$$
  • $\begingroup$ So, there are no further solutions to the given equation ? $\endgroup$ – Peter Mar 20 '18 at 14:55
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    $\begingroup$ indeed.${}{}{}{}$ $\endgroup$ – punctured dusk Mar 20 '18 at 14:56

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