It was also in german IMO selection test (I have a clue where it's from... ) I didn't find the solution in test itself

I'm having a little problem with this stuff, the problem is that I can prove: (1) for all integer n there exist infinitely many a such that t(a*n) != n But I cannot prove (since i have counterexamples ) (2) there exist infinitely many a such that for all integers n : t(a*n) !=n I'm not sure about the bracket : is it p^p-1 or p^(p-1) but i have counterexample for both case (assuming k=p^p-1 or k=p^(p-1)) In fact by putting p=2, i get a=p^(p-1) = 2, and then by putting n=4 I get t(2*4) = 4 And the other way by putting p=2 , i get a=p^p-1=2^2-1=3, and now by putting n=3 I get t(3*3) = 3... I'm really confused... Are you sure the statement of the problem is exactly the thing you typed ? Or is the hint k=p^p-1 wrong ? Or simply I didn't understand the problem ?

The hint is: $a=p^{p-1}$, and $p$ has to be big enough

ok, just choose $a=p^{p-1}$ it is easy to show $p$ must divides $n$, so we get: $d(p^m)d(a')=a'p^{m-p+1}$ from here, use the fact that $d(a')<a/2+1$ with equality only for $a=2$, and do some inequalities based on induction on $m$, to show rhs is always bigger! As for the other problem posted here: iff $n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ put $a=p_1^{p_1^{a_1}-a_1-1}p_2^{p_2^{a_2}-a_2-1}...p_k^{p_1^{a_k}-a_k-1}$ A simple induction in $a_k$ shows the exponents are indeed nonegative integers, so thats it. Finally for the problem 2, the statement must be wrong, taking $n=1$! we get $a=1$ but it doesnt works! Iff 1 is not allowed, taking primes wich doesnt divide $a$ we get $d(a)=1$!

Let $a=p^{p-1}$, $p>2$. So suppose $p$ does not divide $n$ and $\tau(an) = p\tau(n)=n$, which is contradiction. So let $n=p^{a_1}p_2^{a_2}...p_k^{a_k}$ hence $\tau(an)=(a_1+p)(a_2+1)....(a_k+1) = p^{a_1}p_2^{a_2}...p_k^{a_k}$ We can easily prove from induction that $p^{m} > m+1$, $p>2$ and $p^m \geq m+p$ for $m>1$ , $p>2$. So LHS<RHS if $a_1>1$, so $a_1=1$. If $k=2$ and $p_2=2$, then we have $1+p=p$, which is false. Otherwise let $p_i \geq 3$ be a prime factor of $n$. Consider $\frac{p^m}{m+1}$ for postive integers $m \geq 1$, $p \geq 2$ If $p>q$ then $\frac{p^m}{m+1} > \frac{q^m}{m+1}$ If $j>m$ then $\frac{p^j}{j+1} > \frac{p^m}{m+1}$(this can be proven from induction, too lazy to write it out but it's straight forward) Thus $\frac{p(p_i^{a_i})}{(p+1)(a_i+1)} > \frac{p(3^1}{(p+1)(2)} >1$, since $p>2$ so $3p > 2p+2$. Hence $p*p_i^{a_i} < (p+1)(a_i+1)$, so $\tau(an)=(1+p)(a_2+1)....(a_k+1) < pp_2^{a_2}...p_k^{a_k}$, hence there's no positive integer solution to the given equation when $a=p^{p-1}$ for all $p>2$, which is qed.

Its well known that $T(n) <= 2\sqrt{n}$ so taking $a = p^{p-1}$ we assume there is some x solving the equation and say $x = p^yz \rightarrow (p+y)T(z) = p^yz \rightarrow 2\sqrt{z}(p+y) \ge p^yz \rightarrow 2*(p+y) \ge p^y\sqrt{z}$ and this last inequality is clearly false for large enough $p$ is $y \ge 2$ and if $y = 1$ then $z \le 4$ and brute forcing these values of $z$ shows failure as desired.

Let $n=\prod_p p^{\alpha_p}$ and $a=\prod_p p^{\beta_p}$, where $\alpha_p=v_p(n)$, $\beta_p=v_p(a)$ for each prime $p$. Let $Q$ be some non-empty set of primes of the form $4k-1$. We have $$\frac{\tau(an)}{n}=\prod_p \frac{\alpha_p+\beta_p+1}{p^{\alpha_p}}.\qquad (1)$$ Also, the function $\alpha\mapsto \frac{\alpha+\beta+1}{p^{\alpha}}$ is non-increasing over the non-negative integers $(*)$, as its value at $\alpha-1$ divided by its value at $\alpha$ is at least $1$. Let $\beta_q=q-1$ for $q\in Q$ and $\beta_q=0$ otherwise. If $\tau(an)=n$, then we cannot have $\alpha_q=0$ for and $q\in Q$, because then the numerator of one of the fractions in $(1)$ is $q$, but none of the denominators are divisible by $q$, so the product cannot be $1$. So we have $\alpha_q\ge 1$ for each $q\in Q$. Let there be $x$ of them with $\alpha_q=1$ and $y$ of them with $\alpha_q\ge 2$. Note that by $(*)$, if $\alpha_q\ge 2$, then $$\frac{\alpha_q+\beta_q+1}{p^{\alpha_q}}\le \frac{q+2}{q^2}<\frac{2q}{q^2}<1.$$ Also, if $\alpha_q=1$, then $$\frac{\alpha_q+\beta_q+1}{p^{\alpha_q}}=\frac{q+1}{q}< 2.$$ Now note that here, $4\mid q+1$ for $q\in Q$. So in product $(1)$, we have $x+y=|Q|$ factors of $4$ in the numerator, which need be compensated by $\alpha_2\ge 2|Q|$. We get from $(*)$ that $$\frac{\alpha_2+\beta_2+1}{2^{\alpha_2}}\le \frac{2|Q|+1}{4^|Q|}< (1/2)^{|Q|},$$ where the last inequality is due to $2|Q|+1<2^{|Q|}$, assuming $|Q|\ge 3$. Also, for $p\notin Q$, the corresponding fraction is at most $1$. Hence our product $(1)$ is less than $2^x\cdot (1/2)^{|Q|}<1$, and thus cannot be equal to $1$. This proves that for any such set $Q$ with $|Q|\ge 3$, we can find a new $a$ which works. There are infinitely many such sets as there cannot be finitely many primes of the form $4k-1$ (if there were only the ones $q_1,q_2,\dots,q_t$, then the number $4q_1q_2\dots q_t-1$ would have a prime factor $4k-1$, yet it is not divisible by any of the $q_i$, contradiction).

Oops. Here is a slight overkill. Oops. The "slight overkill" is also wrong. We begin with a claim: There is a constant $C>0$ so that\[ \tau(n) < C\sqrt[3]{n} \]for all $n$. Indeed, notice that for $p = 2,3,5,7$, there is a constant $c_p>0$ so that $(\alpha+1)^3 \leq c_pp^{\alpha}$ for all integer $\alpha \geq 0$; further note that if we set $c_p = 1$ for primes $p>8$ the inequality also holds. So, let $C = \sqrt[3]{c_2c_3c_5c_7}+1$. Then, write $n = \prod p_i^{\alpha_i}$, and then we want \[ \prod (\alpha_i+1)^3 < C^3 \prod p_i^{\alpha_i}. \]But by assumption, \[ \prod (\alpha_i+1)^3 \leq \prod c_{p_i}{p_i}^{\alpha_i} < C^3 \prod p_i^{\alpha_i}, \]since $c_p = 1$ for $p > 8$. Hence the claim is true. Notice that if $n$ is a solution to $\tau(an) = n$, then \[ n = \tau(an) < C\sqrt[3]{an}, \]so $n^3 < C^3an$, or $n< C^{\frac 32} \sqrt a$. Now, suppose by contradiction that only finitely many $a$ exist; then, for $a>A$, there is a solution $n_a$ to \[ \tau (an_a) = n_a. \]Pick some $N\in \mathbb N$, and consider \[ n_{A+1}, \ldots, n_{A+N}. \]We have $n_{A+i} < C^{\frac 32} \sqrt{A+i} \leq C^{\frac 32} \sqrt{A+N}$. However, notice that the $n_{A+i}$ are all distinct, so there must be at least $N$ positive integers $x$ (namely our $n_{A+i}$) satisfying $x < C^{\frac 32} \sqrt{A+N}$; this implies that $N < C^{\frac 32} \sqrt{A+N}$, which is clearly a contradiction for $N$ big. $\blacksquare$

Here's my solution.

MathStudent2002 wrote: ...Now, suppose by contradiction that only finitely many $a$ exist; then, for $a>A$, there is a solution $n_a$ to \[ \tau (an_a) = n_a. \]Pick some $N\in \mathbb N$, and consider \[ n_{A+1}, \ldots, n_{A+N}. \]We have $n_{A+i} < C^{\frac 32} \sqrt{A+i} \leq C^{\frac 32} \sqrt{A+N}$. However, notice that the $n_{A+i}$ are all distinct,... Why are $n_{A+i}$ all distinct? No! Not so simple, but I like your approach. A natural one. I also like very much to solve NT problems without using NT at all (doesn't mean I don't like NT). So, $n_a$ may not be distinct, but it could be fixed, I think. When $a$ runs over some interval of length, say $n$, $n_a$ runs over an interval of length $\sqrt{n}$, thus for some $\sqrt{n}$ amongst $a$'s , $n_a$ would take the same value. Could we get a contradiction out of it?

Why dont we use $p-3$ instead of $p-1$ ? I think it solves problem immediately. ($p>5$)

FISHMJ25 wrote: Why dont we use $p-3$ instead of $p-1$ ? I think it solves problem immediately. ($p>5$) I don't think that immediately solves the problem, because if $p-2$ is prime, we can let $n=(p-2)\cdot 2^3$. Solution: Let $n=p_1^{k_1}p_2^{k_2}\dots p_m^{k_m}$ for distinct primes $p_i$ and positive $k_i$. We see that $k_i+1\le p_i^{k_i}$ $(*)$ for all primes $p_i$ and positive integers $k_i$ with equality only when $(p_i,k_i)=(2,1)$. Thus, $\tau(n) = (k_1+1)\dots(k_m+1)\le p_1^{k_1}\dots p_m^{k_m}=n$ with equality if and only if $n=2$. Consider $a=p^{p-1}$ for prime $p\ge 5$. I claim that there are no solutions to $\tau(an)=n$. Suppose, for the sake of contradiction, that there exists a solution $n$ to $\tau(an)=n$. If $p\nmid n$, then $p\mid \tau(an)=n$, a contradiction. So, we can let $n=p^{k_1}\cdot p_2^{k_2}\dots p_m^{k_m}$ and $\tau(p^{p-1}\cdot n)=(p+k_1)(k_2+1)\dots(k_m+1)$. If $k_1\ge 2$, then $p+k_1<p^{k_1}$ and $(k_2+1)\dots(k_m+1)\le p_2^{k_2}\dots p_m^{k_m}$ by $(*)$, so the RHS would be greater than the LHS. So, we must have $k_1=1$, and our equation becomes $$(p+1)(k_2+1)\dots(k_m+1)=pp_2^{k_2}\dots p_m^{k_m}.$$ This means that there exists some $k_j$ such that $k_j+1=qp$ for a positive integer $q$. $$(p+1)(k_2+1)\dots qp\dots (k_m+1)=pp_2^{k_2}\dots p_m^{k_m}$$$$(p+1)(k_2+1)\dots q\dots (k_m+1)=p_2^{k_2}\dots p_m^{k_m}$$ Consider $q(p+1)=(k_j+1)\cdot \frac{p+1}{p}$. For $p>2$, we have that $(k_j+1)\cdot \frac{p+1}{p}<p_j^{k_j}$ for all primes $p_j\ge 3$ and positive integers $k_j$. Combined with $(*)$, if $p_j\ge 3$, there is a contradiction, as the RHS would be greater than the LHS. So, we need $p_j=2$ and $(k_j+1)\cdot \frac{p+1}{p}\ge 2^{k_j}$. Remembering that $p\ge 5$, this leads to $k_j=1$ and $\frac{p+1}{p}=1$, a contradiction. Thus, for all $a=p^{p-1}$ for prime $p\ge 5$, there are no solutions to $\tau(an)=n$.

what was the motivation behind taking a^p-1.

Fermat_Theorem wrote: Then $\tau(an)=\tau(p^{p+\alpha-1}q)=n= p^\alpha q $$\newline$ $ \Rightarrow \tau(p^{p+\alpha-1})\tau(q)=p^\alpha q$$\newline$ $\Rightarrow p^{p+\alpha}\tau(q)=p^\alpha q \Rightarrow p^p\tau(q)=q \Rightarrow p|q$, which contradicts the fact that $(p,q)=1$. I think there's a mistake probably which can be fixed tho

Solution from Twitch Solves ISL: The construction is quite specific: we claim that \[ a = p^{p-1} \]works for sufficiently large primes $p > 100$. Indeed, assume for contradiction $\tau(an) = n$ for some integer $n$. Then: If $p \nmid n$, then $\nu_p(an) = p-1$ so $p \mid \tau(an)$, a contradiction. Thus we must have $p \mid n$. Now let $n = p^e t$ for some $e > 0$. Then \[ \underbrace{(e+p) \cdot \tau(t)}_{=\tau(an)} = \underbrace{p^e \cdot t}_{=n}. \]Now we split into two further cases. If $e > 1$ then $p^e > p+e$ and $t \ge \tau(t)$, give a size contradiction together. If $e = 1$ we get $\tau(t) = \frac{p}{p+1} t \ge \frac{101}{102} t$. However, it's known (and easy to show) that $\tau(t) < 2\sqrt t$ for all integers $t$. Hence we get $\frac{101}{102} t < 2 \sqrt t$, so $t < 5$. But the equation $\tau(t) = \frac{p}{p+1} t$ gives no solution for $t = 1, 2$, $p = 2$ for $t = 3$, and $p = 3$ for $t = 4$. So when $p > 100$ the situation $t < 5$ cannot occur either. Having checked everything, the proof is complete.

This was more algebra than number theory. Note (Since I was a bit lazy in writing everything down): All the inequalities I have left unproved are easily solved by a straightforward induction with suitable base cases. We show that for each prime $p\geq 5$ and $a=p^{p-1}$, the equation $\tau(an)=n$ has no integer solution $n$. This would solve the problem. Call a positive integer $n$ rocking, if it is a solution to $\tau(ax)=x$ for $a$ of the above mentioned form. First, if $\gcd(p,n)=1$, then $n$ is not rocking since otherwise $\tau(an)=\tau(a)\tau(n)=p\cdot\tau(n)=n$ and so $p|n$ which is impossible. Thus we may write $n=p^ep_1^{e_1}\prod_{i=2}^m p_i^{e_i}$ for distinct primes $p,p_1=2,p_2,\dots,p_m$ and positive integers $e,e_2,\dots,e_m$ and non-negative $e_1$. If $n$ is rocking, we have $\tau(an)=(e+p)\prod_{i=1}^m(e_i+1)=p^e\prod_{i=1}^m p_i^{e_i}$. Now we can easily show that for $p_i\geq 2$ and $e_i\geq 0$, we have $p_i^{e_i}\geq e_i+1$. So $\prod_{i=1}^m p_i^{e_i}\geq \prod_{i=1}^m(e_i+1)$. So we must have $p^e\leq p+e$. Now since $p\geq 5$, so for $e>1$ we can again show that $p^e>p+e$. Thus we must have $e=1$. This means $(p+1)\prod_{i=1}^m(e_i+1)=p\prod_{i=1}^m p_i^{e_i}$. Now for $p_i>2$, we have $p_i^{e_i}\geq 3^{e_i}>\frac{4}{3}(e_i+1)$ for all positive integers $e_i$. We also have $2^{e_1}\geq e_1+1$ for all non-negative integers $e_1$. Hence we have $\left(1+\frac{1}{p}\right)\cdot1\geq \left(1+\frac{1}{p}\right)\left(\frac{e_1+1}{2^{e_1}}\right)=\frac{\prod_{i=2}^m p_i^{e_i}}{\prod_{i=2}^m(e_i+1)}\geq \left(\frac{4}{3}\right)^{m-1}$. But for $p\geq 5$ this is only possible if $m-1=0$ or $m=1$. Thus $n=p\cdot 2^{e_1}$. This gives $(p+1)(e_1+1)=p\cdot2^{e_1}$. This means that $e_1>0$. Also, $e_1=1$ doesn't work. Now if $e_1>1$ then $2^{e_1}>\frac{5}{4}(e_1+1)$, so we get $1+\frac{1}{p}=\frac{2^{e_1}}{e_1+1}>\frac{5}{4}=1+\frac{1}{4}$ which contradicts the fact that $p\geq 5$. Hence $n$ cannot be rocking, and we are done. $\square$

I think that this solution is new. Claim 1. There exist no solutions to the equation \[ 6 \tau(n) = n . \]Proof Obviously $6$ must divide $n$, so write the prime factoirsation $n = 2^{1 +e_1} \cdot 3^{1+e_2} + p_3^{e_3} \cdots p_k^{e_k}. $ Then \[ \frac{e_1+2}{2^{e_1}} \cdot \frac{e_2 + 2}{3^{e_2}} = \prod_{i=3}^k \frac{p_i^{e_i}}{e_i+1}. \]Observe that $x \to \frac{x+2}{2^x}$ and $\frac{x+2}{3^x}$ are both strictly decreasing functions, so the lefthand side is at most 4. Now we place some lower bounds on the righthand side. Note that the function $\frac{p^x}{x+1}$ is increasing for any prime $p > 3$. If $n$ is divisible by a prime $p \geq 11$, then $\frac{p^{e_i}}{e_i + 1} \geq p/2 > 4$. Thus there are no solutions if $n$ is divisible by a prime $p \geq 11$. If $n$ is divisible by $7^2$, then the righthand side is at least $\frac{49}{3} > 4$, so there are no solutions with $n$ divisible by 49. If $n$ is divisible by $5^2$, then the righthand side is at least $\frac{25}{3} > 4$, so there are no solutions with $n$ divisible by 25. If $n$ is divisible by $5 \cdot 7$, then the rightside is at least $\frac{5 \cdot 7}{4} > 4$, so there are no solutions here. If $n = 2^{a+1} \cdot 3^{b+1} \cdot 5$, then \[ \frac{a+2}{2^a} \cdot \frac{b+2}{3^b} = \frac{5}{2}. \]This has no solutions for size reasons when $a \geq 2$ or $b \geq 1$, so we just have to manually check $n=60$. if $n = 2^{a+1} \cdot 3^{b+1} \cdot 7$, then \[ \frac{a+2}{2^a} \cdot \frac{b+2}{3^b} = \frac{7}{2} \]which likewise has no solutions when $a \geq 2$ or $b \geq 1$ for size reasons. Thus we check $n = 84$, and we are done. This establishes the claim. Claim 2. Let $C$ be a constant. Then \[ C\tau(n) < n \]for sufficiently large $n$. Proof. This is more dumb bounding. Note that \[ \frac{n}{\tau(n)} = \prod_{p \mid n} \frac{ p^{\nu_p(n)}}{\nu_p(n) + 1} \geq \frac{ q^{\nu_p(n)}}{\nu_q(n) + 1} \]for any fixed prime $q$. Obviously, $x \to \frac{q^x}{x+1}$ is unbounded and increasing. Thus $\nu_q(n)$ is bounded for all primes $q$. Hence $n$ is bounded. Now let $N$ be the integer such that $\frac{7}{2} \tau(n) < n$ for all $n \geq N$ and choose a prime $p \geq N$. Then pick $a = p^5$. If $p$ does not divide $n$, then $\tau(p^5n) = 6\tau(n)$, which has no solutions by Claim 1. Thus assume $n$ is divisible by $p$ and write $n = p^k m$ where $m$ is not divisibly by $p$. Then \[ \tau(p^5n) = \tau(n) \left(1 + \frac{5}{k+1}\right) \leq \frac{7}{2} \tau(n) \]Since $n \geq p > N$, we have that this is less than $n$, so we are done.

We claim that $a=p^{p-1}$ works, where $p\ge 5$ is a prime number. Assume that there exist $n=\tau(an)$. If $p$ doesn't divided $n$, then $n=\tau(p^{p-1}n)=p\tau(n)$, contradiction. So let $n=p^km$, we get $$0=n-\tau(p^{p-1}n)=p^{k}m-(p+k)\tau(m)=\sum_{d|m} p^k\varphi(d) - \sum_{d|m} (p+k)$$If $k> 1$, then $p^k\varphi (d) -(p+k) > 0$ contradiction, so consider that $k=1$, that means $0=\sum_{d|m} p(\varphi(d) - 1)-1$ . $n \neq 1, 2$ of course, and for $d> 2$, we have $p(\varphi(d)-1)-1\ge p-1> 3$, only when $d=1, 2$, $p(\varphi(d)-1)-1= -1$, hence the summation is positive, contradiction. So we're done.

Yes, I agree.

We claim that $a=p^{p-1}$ works for all primes $p$ sufficiently large. Lemma: $p \mid n$. Assume not. Then we have that $$\tau (p^{p-1}n) = p \cdot \tau (n) = n,$$contradiction. Let $n=p^km$ for some $m$ such that $p \nmid m$. Then $$\tau (p^{p+k-1}m) = (p+k) \tau(m) = p^k m$$. Note that when $k=1$, $v_p(p)=1$ and $v_p(p+1)=0$, and for $k>1$, $p^k>p+k$ and $p^k$ has only $p$'s in its prime factorization, we get that $v_p(p^k)>v_p(p+k) \implies p \mid m$. This shows that $m$ is sufficiently large. Using $\tau (m) \le 2\sqrt m \le \frac{m}{2}$, because $m$ is sufficiently large, we get that $$p^km=(p+k) \tau(m) < \frac{m(p+k)}{2} \implies 2p^k<p+k,$$an obvious contradiction.

Note our problem is equivalent to showing that the sequence $$ \frac{1}{\tau(1)}, \frac{2}{\tau(2)}, \frac{3}{\tau(3)}, \ldots $$does not contain infinitely many positive integers, since if $$ \tau(an) = n \implies \frac{an}{\tau(an)} = a $$Conversely, if $$ \frac{m}{\tau(m)} = a \implies m = an ~~ \text{for some } n \in \mathbb Z_{>0} $$Fix prime $p \ge 5$. We claim numbers of the form $p^{p-1}$ are not in the sequence. FTSOC $$ \frac{k}{\tau(k)} = p^{p-1} $$Then $\nu_p(k) \ge p-1$. Note $\nu_p(k)$ cannot equal $p-1$ (otherwise $\nu(p)$ of LHS would equal $p-2$, as $p \mid (p-1) + 1$, so a factor of $p$ would be in denominator also). Thus $\nu_p(k) \ge p$. Now if $e = \nu_p(k) \ge p+1$, then $$ \frac{k}{\tau(k)} \ge \frac{p^e}{e+1} > p^{p-1} $$which gives an contradiction. Thus $\nu_p(k) = p$. But then for some prime $q$, $\nu_q(k)+1 \mid p$, more particularly $f = \nu_p(k) \ge p-1$. Thus, $$ \frac{k}{\tau(k)} \ge \frac{p^p}{p+1} \cdot \frac{q^f }{f+1} \ge p^{p-1} \cdot \frac{q^{p-1}}{p+1} \ge p^{p-1} \cdot \frac{2^{p-1}}{p+1} > p^{p-1} $$which gives a contradiction. $\blacksquare$

I have a potential patch: Let $p$ be sufficiently large. From #31, we deduce that $p\mid n$. Let $m=\frac n{p^{v_p(n)}}$. First, note that: $$p^{v_p(n)}m=n=\tau(p^{p-1}n)=\tau(p^{v_p(n)+p-1}m)=(v_p(n)+p-1)\tau(m).$$Assume that $m\ge4$, then: $$p^{v_p(n)}m=(v_p(n)+p-1)\tau(m)\le2(v_p(n)+p-1)\sqrt m\le\frac m2(v_p(n)+p-1)$$so $2p^{v_p(n)}\le v_p(n)+p-1$. By Bernoulli's inequality, $2+(2p-2)v_p(n)\le2v_p(n)+2p-2$ which rewrites as $(2p-4)v_p(n)\le2p-4$. If $v_p(n)>1$ then we have $2p-4>2p-4$, a contradiction. So $v_p(n)$ is exactly $1$. But then $2p\le p$, so no $n$ exists. We can similarly eliminate $m=1,2,3$.

We claim that for $a=5^4$ there is no solution. Assume towards a contradiction there was. Let $n=5^kX$ where $5$ does not divide $X$. Then we have $$\tau(5^{k+4}X)=5^kX$$$$(k+5)\tau(X)=5^kX$$We know $k\neq 0$ since otherwise the LHS would have a factor of 5 and the RHS wouldn't. Also, since $X\geq \tau(X)$, $$k+5\geq 5^k$$So, $0<k\leq 1$, so $k=1$. So, $6\tau(X)=5X$. So, $\frac56=\frac{\tau(X)}{X}$. Since only $X$ and the numbers less than or equal to $\frac{X}2$ can be divisors of $X$, we know $\tau(X)\leq 1+\frac{X}2$. So, $$\frac{\tau(X)}{X}\leq \frac1{X}+\frac12$$$$\frac56\leq\frac1{X}+\frac12$$$$\frac13\leq\frac1{X}$$$$3\geq X$$However, $X$ must be a multiple of $6$, a contradiction. Now we will show there are infinitely many other numbers for which there is no solution. Let $a=p^{5^4-1}$ for a large prime $p$. Assume towards a contradiction $n$ is a positive integer such that $\tau(p^{5^4-1}n)=n$. Let $n=p^aX$ where $p$ doesn't divide $X$. Then, $$(5^4+a)\tau(X)=p^aX$$So, $$5^4+a\geq p^a$$For sufficiently large $p$, this implies $a=0$. So, plugging in $a=0$, we get $$5^4\tau(X)=X$$Let $k=\tau(X)$. Then, $$5^4\tau(5^4k)=5^4k$$$$\tau(5^4k)=k$$a contradiction.

We claim $\boxed{a=p^{p-1}}$ for any prime $p>547$ satisfies that $\tau(an)=n$ has no positive integer solution $n$. Suppose there existed $n$ with $\tau(p^{p-1}n) = n$. Claim: $p$ divides $n$. Proof: Suppose not. Then $\nu_p(p^{p-1} n )=p-1$, so $p\mid \tau(p^{p-1}n)=n$, contradiction. $\square$ Let $n=c\cdot p^k$, where $p\nmid c$ and $k=\nu_p(n)>0$. So we have $\tau(c\cdot p^{k+p-1} ) = c\cdot p^k$ Thus, $\tau(c) \cdot (k+p) = c\cdot p^k$. If $k>1$, we have $c\ge \tau(c) $ and $p^k > p+k$, so $c\cdot p^k > \tau(c)\cdot (p+k)$, contradiction. So $k=1$. Thus, $\tau(c) \cdot (p+1) = c\cdot p$, so $\tau(c) = \frac{p}{p+1}\cdot c$. Claim: $\tau(c) \le 2\sqrt{c}$. Proof: If not, then the number of unordered factor pairs multiplying up to $n$ is greater than $\sqrt{c}$. However, the least number in the factor pair has to be at most $\sqrt{c}$, a contradiction because each positive integer is in at most one ordered factor pair of $c$. $\square$ So we have $c\cdot \frac{p}{p+1} \le 2\sqrt{c}$. Squaring both sides gives \[c^2\cdot \left(\frac{p}{p+1}\right)^2 \le 4c\implies c\cdot \left(\frac{p}{p+1}\right)^2\le 4 \] However, since $p>187$, we have $4> c\cdot \left(\frac{187}{548}\right)^2= \frac{299209c}{300304}\implies c<\frac{1201216}{299209}<5$ Case 1: $c=4$. Then $3=\frac{4p}{p+1}$. This implies $3\mid p$, not possible as $p>187$. Case 2: $c=3$. Then $2=\frac{3p}{p+1}$, which implies $p$ is even, not possible as $p>187$. Case 3: $c=2$. Then $2=\frac{2p}{p+1}$, so $\frac{p}{p+1} = 1$, not possible. Case 4: $c=1$. Then $\frac{p}{p+1} = 1 $, not possible. We have exhausted all cases, so we are done.

Same as above solutions; posting for storage. We claim $a=p^{p-1}$ works for $p>1234.$ Suppose FTSOC the equation has solutions. We claim $p\mid n$; assume the contrary. Then, $$n=\tau(p^{p-1}n)=\tau(p^{p-1})\tau(n)=p\tau(n),$$a contradiction. Let $n=p^{\alpha}m$ where $\alpha>0$ and $p\nmid m.$ If $\alpha>1,$ clearly $p+\alpha<p^{\alpha}.$ Since $\tau(m)\le m,$ we see $$\tau(an)=\tau(p^{p-1+\alpha}m)=(p+\alpha)\tau(m)<p^{\alpha}m=n,$$which is absurd. Thus, $\alpha=1.$ It is well known that $\tau(m)<2\sqrt{m},$ so if $m\ge 5,$ we have $$m>4(1+1/p)^2\implies n=pm>(p+1)2\sqrt{m}>\tau(m)\tau(p),$$which leads to no solutions. If $m=4,$ then $(p+1)3=4p,$ or $p=3.$ If $m=3,$ then $(p+1)2=3p,$ or $p=2.$ If $m=2,$ then $(p+1)2=2p,$ which yields no solutions. If $m=1,$ then $p+1=p,$ which is absurd. Since $p>1234,$ we have no solutions in any of these cases. $\square$

Let $n=p^{e} \cdot \prod_{i=1}^{j}{p_i^{e_i}}$ where $p_1<p_2<\dots<p_j$ and $p$ is distinct from all the other primes and is odd. Let $a=p^{p-1}$ then $\tau(an)=(e+p)(e_1+1)(e_2+1)\dots (e_n+1).$ If $e=0$ then $p\nmid n$ but then $n=\tau(an)=p\tau(n)$ so there aren't solutions. Now, $p\mid n.$ If $\nu_{p}{n}>2$ then for each $i$ we have $p_i^{e_i}\ge e_i+1$ with equality iff $p_i=2.$ Then, $e>2\implies p^e>p+e$ so we have no solutions. Thus, if there is a solution then $e=1.$ We have $(p+1)(e_1+1)\dots(e_j+1)=p\cdot p_1^{e_1}p_2^{e_2}\dots p_j^{e_j}.$ Obviously we have $p\mid e_i+1$ for some $i$ then $e_i\ge p-1.$ Then, we have $p_i^{p-1}>p+1$ whenever $p>3$ and we use $p_i^{e_i}\ge e_i+1$ for all the other $i$ and we win.

amuthup wrote: We claim that all $a$ of the form $p^{p-1}$ for a prime $p\ge 3$ work. beta wrote: Let $a=p^{p-1}$, $p>2$. For $p = 3$, $p^{p-1} = 9$, and $\tau(9\cdot 12) = 12$. HKIS200543 wrote: Claim 1. There exist no solutions to the equation \[ 6 \tau(n) = n . \] [Nevermind this was already mentioned by someone else earlier in the thread] $n = 72$. (Your proof of the claim fails to check $n = 2^a3^b$.)

Solved with motivation for bashing from my classmates. I claim that $a=p^5q^2$ where $p$ and $q$ are distinct primes greater than $100$ always works, which obviously finishes. Claim. $\gcd(n,a)>1$. Proof. Assume FTSOC that $\gcd(n,a)=1$. Then $d(an)=18d(n)$. So $d(n)=n/18$. If $n>18\cdot 72$, then \[d(n)\le 2\sqrt n <\frac{n}{18},\]so $n\le 18\cdot 72$. $1680$ is a highly composite number greater than $18\cdot 72$, and it has $40$ positive divisors, so $n\le 18\cdot 40$. I tested all positive multiples of $18$ less than or equal to $18\cdot 40$, and none of them worked. $\blacksquare$ Now let $n=p^x q^y m$ where $p\nmid m$ and $q\nmid m$. Then \[p^x q^y m = n = d(pqn) = d\left(p^{x+5}q^{y+2}m\right) = (x+6)(y+3)d(m), \]meaning that \[\frac{x+6}{100^x}\cdot\frac{y+3}{100^y} \ge \frac{(x+6)(y+3)}{p^x q^y} = \frac{m}{d(m)} \ge 1.\]But since $x+y>0$ by the previous claim, the LHS is less than $1$, contradiction. $\blacksquare$

We claim all $a=p^{p-1}$ for prime $p\geq 5$ work. Suppose $n$ is a solution to $\tau(p^{p-1}n)=n$. Suppose $p\nmid n$. Then $p\tau(n)=\tau(p^{p-1}n)=n$, which is a contradiction. So, $p\mid n$ and $n=p^kn'$ for some positive integer $k$ and positive integer $n'$ not divisible by $p$. We now have the equation \[(p+k)\tau(n')=\tau(p^{p-1}p^kn')=\tau(p^{p-1}n)=n=p^kn'\]We now use the bound \[\tau(m)\leq 2\sqrt{m}\]Which can be shown using the fact that factors of $m$ come in pairs, one being $\geq \sqrt{m}$ and one being $\leq \sqrt{m}$. Applying the bound, we get $\frac{p^k}{p+k}n'=\tau(n')\leq 2\sqrt{n'}\implies \frac{p^k}{p+k}\leq \frac{2}{\sqrt{n'}}$. Case 1: $k=1$ Using $(p+1)\tau(n')=pn'$, we get that $p\mid \tau(n')$. So, $n'$ has some divisor of the form $q^{dp-1}$, which is at least $2^4=16$. So $n'\geq 16$ and we have $\frac{1}{2}<\frac{p}{p+1}\leq \frac{2}{\sqrt{n'}}\leq \frac{1}{2}$, which is a contradiction. Case 2: $k>1$ $n'=1$ or $n'=2$ gives $p+k=p^k$, and $n'=3$ gives $p+k=\frac{3}{2}p^k$. Each of these are contradictions since $p^k>p+k$. So, $n'\geq 4$. We then have $1<\frac{p^k}{p+k}\leq \frac{2}{\sqrt{n'}}\leq 1$, a contradiction. So, no such $n$ exist and the infinitely many $a$ of the form $p^{p-1}$ for prime $p\geq 5$ work.

>consider $p^{p-1}$ immediately >do not find a proof >come up with whatever this is I claim that $\boxed{p^{14}}$ does the job for sufficiently large primes $p$. Suppose we prime factorize $n=p^{f}p_1^{f_1}\ldots p_k^{f_k}$ for positive integers $f_i$. Then we need $(f+15)(f_1+1)\ldots(f_k+1)=p^{f}p_1^{f_1}\ldots p_k^{f_k}$. Let $\omega(m)$ denote the number of prime powers greater than $1$ dividing $m$; note that $\omega$ is completely multiplicative. Applying it to both sides of the equation, we get $$\omega(f+15)+\omega(f_1+1)+\cdots+\omega(f_k+1)=f+f_1+\cdots+f_k \implies (f_1-\omega(f_1+1))+\cdots+(f_k-\omega(f_k+1))=\omega(f+15)-f$$ Since $\omega(m) \leq \tau(m)\leq 2\sqrt{m}$, where the last inequality is well-known and easy to prove, it follows that for fixed positive integers $r$, $\omega(m+r)-m$ is bounded above and eventually negative as $m$ varies. Furthermore, we can check (by examining small $m$ manually) that we must always have $m-\omega(m+1) \geq 0$. In particular, this means that, by appealing to the fact that $\omega(f+15)-f$ is bounded above, $f_i-\omega(f_i+1)$ is bounded above as well, hence the $f_i$ are all bounded above. Furthermore, since $f-\omega(f+15)$ should be nonpositive, $f$ is bounded above as well. Hence if $p$ is sufficiently large, we are forced to have $f=0$, else $p$ divides $n$ but $p \nmid (f+15)(f_1+1)\ldots(f_k+1)$ for size reasons. Hence the RHS of the equation just equals $2$, forcing $f_i-\omega(f_i+1) \leq 2$ for all $i$. We may manually verify that for positive integers $m$, $m-\omega(m+1) \leq 2 \implies m \in \{1,2,3\}$, hence $f_i \in \{1,2,3\}$ for all $i$. Now we do casework. Because $15 \mid n$, we should have $\nu_5(n) \in \{1,2,3\}$. On the other hand, if $\nu_5(n) \geq 2$ then we need some prime $q$ to have $\nu_q(n) \equiv 4 \pmod{5}$, but this contradicts the fact that it should be at most $3$ always. Thus $\nu_5(n)=1$. We should also have $\nu_3(n) \in \{1,2,3\}$. We consider these three cases. If $\nu_3(n)=1$, then $\nu_2(n) \in \{2,3\}$, since now $(1+1)(1+1) \mid \tau(an)$. But $\nu_2(n)=2$ would imply $9 \mid \tau(an)$ and $\nu_2(n)=3$ would imply $16 \mid \tau(an)$, so this case is impossible. If $\nu_3(n)=2$, then $\nu_2(n) \in \{1,2,3\}$. But $\nu_2(n)=1$ would imply $4 \mid \tau(an)$: contradiction. On the other hand, if $\nu_2(n) \in \{2,3\}$, we are faced with a problem of having "too many" factors of $2$, so there has to be some prime $q \mid n$ which we haven't considered yet that has odd exponent. But then $q \geq 7$, so $q \nmid 15$ and $q \nmid f_i+1$ due to size reasons: also a contradiction. If $\nu_3(n)=3$, then $\nu_2(n)=3$ is forced since $8 \mid \tau(an)$. But then $32 \mid \tau(an)$, which is again impossible. This finishes the problem. $\blacksquare$ Remark: $a=p^{14}$ is like the 11th thing I considered of this form, but there's actually a reason it works and everything I considered beforehand doesn't. In particular I think $a=p^{3q-1}$ should work as well for $q\geq 5$, if $p$ is taken to be sufficiently large relative to $q$.

We can rewrite this as $\frac{\tau (\alpha n)}{\alpha n} = \frac{1}{\alpha}$. Since the latter is multiplicative over prime powers, it suffices to show there are infinitely many natural numbers whose reciprocals cannot be written in the form $\prod \frac{k_p + 1}{p^{k_p}}$ where the product is taken over distinct primes. We claim all numbers of the from $9p^2$ for sufficiently large $p$ work. Note that for sufficiently large $p$ and $k_p > 2$ , we have $\frac{1}{9p^2} > \frac{k_p + 1}{p^{k_p}}$, note that the rest of the terms in the product are less than $1$ so we cannot have $k_p > 2$, and by $\nu_p$ arguments on the denominator we see that $k_p$ cannot be $0,1$, so $k_p = 2$. Thus we have to show that the product of the terms for the primes other than $p$ is exactly $\frac{1}{9p^2}$ divided by $\frac{3}{p^2}$ which is just $\frac{1}{27}$. To do this, we analyze $k_3$. Clearly, $k_3 < 5$ by size arguments, and $k_3 >2$ by $\nu_3$ arguments, so we analyze $k_3 = 3,4$. In the former case, we require the product of the terms for the primes other than $p,3$ to be $\frac{1}{27}$ divided by $\frac{3 + 1}{3^3}$ which is just $\frac 14$. Clearly if we had $k_p$ above $0$ for any primes above $7$, we would have a product of the other terms less than $\frac 14$, so we can turn our attention to $p = 2,5,7$. By $\nu_2$ arguments, we see that $k_2 > 1$, and by size we see $k_2 < 5$. If $k_2 = 2$, the product for the primes $5,7$ is forced to be $\frac 13$, impossible since the denominator cannot have a factor other than $5,7$, if $k_2 = 3$, the product for the primes $5,7$ is $\frac 12$, impossible since the denominator cannot have a factor other than $5,7$, if $k_2 = 4$ the product for the primes $5,7$ is $\frac 45$, impossible by size. Now we go to the case where $k_3 = 4$, then we have the product for primes other than $p,3$ as $\frac 35$, which requires $k_5 > 0$, but this fails by size.

>consider $p^{p-1}$ immediately >do not find a proof >come up with whatever this is Note that $\tau(n) < Cn^\varepsilon$ holds for all $n$ and some $C$ and fixed $\varepsilon$. We now take \[ a = p^{(q^{q-1}-1)} \]where $p$ is any really large prime and $q$ is a pretty large prime where $q+1$ is divisible by prime $14341$. Note that $d(a) = q^q$ and $\tau(an) \le \tau(a) \cdot \tau(n)$, so taking $p$ such that for all $N > p$, $\tau(a) \cdot CN^\varepsilon < N$ means that any $n$ that satisfies $\tau(an) = n$ can't be divisible by $p$. As such, it follows that $n = \tau(an)$ is divisible by $q^{q-1}$. If$ q^{q} \nmid n$ then this implies that \[ \tau(p^{q^{q-1}-1}q^{q-1}m) = q^{q-1} \cdot q \cdot \tau(m) \ne q^{q-1}m. \]and thus $q^q \mid m$. If $q^{q+1} \mid m$, then \[ \tau(p^{q^{q-1}-1}q^{q+1}m) = q^{q-1} \cdot (q+2) \cdot \tau(m) < q^{q+1}m \]which holds for sufficiently large $q$. Finally, if $q^{q+1} \nmid m$ then we get that $14341 \mid q+1 \mid m$, but then \[ \tau(14341p^{q^{q-1}-1}q^{q}m) = 2 q^{q-1} \cdot (q+1) \cdot \tau(m) < 14341 q^qm \]which holds for sufficiently large $q$. This means no solution exists, contradiction.