Let $p$ be a prime dividing $a$. Also let $v_p(a)=\alpha$ and $v_p(b)=\beta$ and assume without loss of generality that $\alpha > \beta$. Let $a=p^{\alpha} p_1^{\alpha_{1}} ... p_m^{\alpha_{m}}$ and $b=p^{\beta} q_1^{\beta_{1}} ... q_l^{\beta_{l}}$. Then we have $\forall k\in \mathbb{N}$, $\tau((\alpha + k)(\alpha_{1} + 1)...(\alpha_{m} + 1)) = \tau((\beta + k)(\beta_{1} + 1)...(\beta_{l} + 1))$. Now let $P$ be a large prime. Take $k$ such that $\alpha + k = 2^{P-1}$. Then, since $P$ is large, $\beta + k$ has to have a prime divisor $q$ such that $v_q(\beta + k) \equiv -1 \pmod{P}$. But since $q \geq 2$, this means that $\beta + k \geq 2^{P-1} = \alpha + k \implies \alpha \leq \beta$, contradiction with our assumption. Thus $\alpha = \beta$ and since $p$ was an arbitrary prime dividing $a$, we must have $a = b$.

$k$ should've been different. Fixed it now.

Here's another solution that's much less efficient and probably equivalent to the above. If $a\neq b$, then $\nu_q(a)<\nu_q(b)$ for some prime $q$. Let $a'=\nu_q(a)$ and $b'=\nu_q(b)$, and let $n=q^{d-1}$ for this $q$ and some variable integer $d>1$. Then since $\tau$ is multiplicative and $\nu_p(an)=a'+d-1$, $$\tau(an)=\tau\left(\frac{an}{q^{a'+d-1}}\right)\tau(q^{a'+d-1})=\underbrace{\tau\left(\frac{an}{q^{a'+d-1}}\right)}_{\text{Call this }K}\times (a'+d).$$Similarly, $\tau(bn)=L\times (b'+d)$, where $L=\tau\left(\frac{an}{q^{b'+d-1}}\right)$. Now note that $K$ and $L$ are constants (i.e. independent of $d$). So if we let $c=b'-a'>0$ and $A=a'+d$, we see that $\tau(\tau(an))=\tau(\tau(bn))$ is equivalent to $$ \tau(KA) = \tau(L(A+c)),\qquad (\dagger)$$where $A$ can attain any sufficiently large value by varying $d$ as needed. Let $p_1,p_2$ be primes larger than $100cL$, and set $A$ to be a prime larger than $K$ and congruent to $-c\pmod{p_1p_2}$ (which we can do by Dirichlet). Then observe that $\tau(L(A+c))\geq 3\tau(L)$, since for each divisor $d$ of $L$, $p_1d$ and $p_2d$ are also divisors of $L(A+c)$ but not $L$. As $A$ is a prime coprime to $K$, we then have $$2\tau(K)=\tau(A)\tau(K)=\tau(KA)\stackrel{(\dagger)}{=}\tau(L(A+c)) \geq 3\tau(L).$$Now consider what happens if we instead set $p$ to be a prime larger than $100cL$, and $A+c$ to be a prime $c\pmod p$ (which we can again do by Dirichlet). For each divisor $d$ of $K$, $pd$ is a divisor of $KA$ but not $K$, so $\tau(KA)\geq 2\tau(K)$, and we similarly establish $$2\tau(L)=\tau(L)\tau(A+c)\tau(L(A+c))\stackrel{(\dagger)}{=}\tau(KA)\geq 2\tau(K).$$Combining, we have $2\tau(L)\geq 2\tau(K)\geq 3\tau(L)$, our desired contradiction. Thus we must have had $a=b$.

Suppose that $\nu_p(a)\neq\nu_p(b)$ for some prime $p$ with $a>b$. Suppose $P,Q$ are primes greater than $\tau(a),\tau(b)$ and note that if we let $n=p^{P^{Q-1}-a}$ for $k\geq a+1$, we must have \[P^{Q-1}\mid\mid\tau(an)\implies Q\mid\tau(\tau(an))=\tau(\tau(bn))\]implying that as $Q>\tau(b)$ and $\tau(x)<x$, we must have that there is a prime $P'$ such that $P'^{Q-1}=P^{Q-1}-a+b$, which is impossible as $P,Q$ are arbitrary. Thus $a=b$ for all primes and we are done.

Solution with neel02 : Lemma: Given any non-zero integer $a$.There exists a prime number $p$ and a integer $m$ such that $p^{2m}+a$ has a prime divisior $q$ such that $v_q(p^2m+a)$ is odd.

Lemma: Given any $r\in \mathbb N$ and $r$ integers $a_1,a_2,\dots,a_r$ such that atleast one of them is non-zero.Then we can find distinct primes $p_1,p_2\dots,p_r$ and positive integers $m_1,m_2'\dots,m_r$ such that the number $$(p_1^{2m_1}+a_1)(p_2^{2m_2}+a_1)\dots(p_r^{2m_r}+a_r)$$has a prime divisiorb $q>3$ such that $$v_q((p_1^{2m_1}+a_1)(p_2^{2m_2}+a_1)\dots(p_r^{2m_r}+a_r))$$is odd.In particular the number is not a perfect square.

Now coming back to the main problem,assume that $$a=p_1^{g_1}p_2^{g_2}\dots p_r{g_r}$$and $$b=p_1^{h_1}p_2^{h_2}\dots p_r{h_r}$$Assume FTSOC,$a\ne b$.Therefore,there is some $i$ for which $g_i\ne h_i$ . Take $n=p_1^{x_1}p_2^{x_2}\dots p_r^{x_r}$ where $x_i=q_i^{c_i}-g_i-1$ where primes $q_i$ and powers will be determined latter. Then \begin{align*} \tau(\tau(an))&=\tau(q_1^{c_1}q_2^{c_2}\dots q_r^{c_r})\\ &=(c_1+1)(c_2+1)\dots(c_r+1) \end{align*}So for all even $c_i$ $\tau(\tau(an))$ is odd. Now observe that, \begin{align*} \tau(\tau(bn))\\ &=\tau((q_1^{c_1}+h_1-g_1)(q_2^{c_2}+h_2-g_2)\dots(q_r^{c_r}+h_r-g_r))\\ \end{align*} Since,there exists at least one $h_k-g_k\ne 0$,by the previous lemma,there exists distincts primes $q_1,q_2\dots q_r$ and even $c_1,c_2\dots c_r$ such that $(q_1^{c_1}+h_1-g_1)(q_2^{c_2}+h_2-g_2)\dots(q_r^{c_r}+h_r-g_r)$ is not a perfect square.So for that particular $n$ \begin{align*} &\tau(\tau(bn))\\ &=\tau((q_1^{c_1}+h_1-g_1)(q_2^{c_2}+h_2-g_2)\dots(q_r^{c_r}+h_r-g_r))\\ &=\text{even} \end{align*} Hence,$\tau(\tau(an))\ne \tau(\tau(bn))$.Contradiction.. $\blacksquare$

Is there any forum for the problem collection of Mock TMO?

here $~~~~~~~~$

hakN wrote: Let $p$ be a prime dividing $a$. Also let $v_p(a)=\alpha$ and $v_p(b)=\beta$ and assume without loss of generality that $\alpha > \beta$. Let $a=p^{\alpha} p_1^{\alpha_{1}} ... p_k^{\alpha_{k}}$ and $b=p^{\beta} q_1^{\beta_{1}} ... q_l^{\beta_{l}}$. Then we have $\forall k\in \mathbb{N}$, $\tau((\alpha + k)(\alpha_{1} + 1)...(\alpha_{k} + 1)) = \tau((\beta + k)(\beta_{1} + 1)...(\beta_{l} + 1))$. Now let $Q$ be a fixed large prime and let $P$ be a large prime. Now take $k$ such that $\alpha + k = Q^{P-1}$, then we must have that $\beta + k = x^{P-1}$ where $x$ is an integer for all large primes $P$. But this is impossible since $\sqrt[P-1]{Q^{P-1} + (\beta - \alpha)} \notin \mathbb{N}$ for very large primes $P$ since $\beta - \alpha$ is a constant and not equal to $0$. So this implies that $\alpha = \beta$ and since $p$ was an arbitrary prime dividing $a$, we must have $a=b$. I think u can't choose $k$ , as $a$ is given fixed in the question..

We will prove that $\nu_p(a)=\nu_p(b)$ for every prime $p$. Assume for contradiction $\nu_p(a)\ne \nu_p(b)$. WLOG, $\nu_p(a)>\nu_p(b)$. Let $r$ be a large prime. We construct $n$ such that $r\mid d(d(an))$ but $r\nmid d(d(bn))$. Let $a'=\prod\limits_{ \nu_q(a) \text{ odd}} q$. We select $n=p^{2^{r-1}-1-\nu_p(a)} a'$. Then $\nu_2(d(an)) = \sum\limits_{q\mid an} \nu_2(\nu_q(an)+1)$ Note $\nu_2(\nu_p(an)+1)=\nu_2(2^{r-1})=r-1$ and $\nu_2(\nu_{p_1}(an)+1)$ for $p_1\ne p$ is 0 because $\nu_{p_1}(an)$ is even. It remains to show $r\nmid d(d(bn))$. $d(bn)=\prod\limits_{q\mid bn} (\nu_q(bn)+1)$ To make sure $r\mid d(d(bn))$, we require $d(bn)=\prod\limits_{q\mid bn} (\nu_q(bn)+1)$ to be divisible by $p_3^{kr-1}$ but not $p_3^{kr}$ for some prime $p_3$ and $k\in \mathbb{N}$. We have $\nu_{p_3} (d(bn))=\sum\limits_{q\mid bn} \nu_{p_3}(\nu_q(bn)+1)$ If $q\ne p$, $ \nu_{p_3}(\nu_q(bn)+1)=\nu_{p_3}(\nu_q(ba')+1)$. Note $\nu_q(ba')+1$ is bounded, so $\sum\limits_{q\mid bn} \nu_{p_3}(\nu_q(ba')+1)$ is bounded. Let this value be $C$ (with respect to $p_3$) We need $\nu_{p_3}(\nu_p(bn)+1)+C=kr-1$ We know $\nu_p(bn)+1=2^{r-1} - \nu_p(a)+\nu_p(b)=2^{r-1} - D$ Therefore, for all primes $r$, there exists $p_3$ such that $C+\nu_{p_3}(2^{r-1}-D)$ is $-1$ mod $r$. I contend $C+\nu_{p_3}(2^{r-1}-D)<r-1$ If $p_3=2$ this is simply $C+\nu_2(D)$ which is bounded, but $r$ can be as large as we want. If $p_3\ge 3$ then $C+\nu_{p_3}(2^{r-1}-D)<C+\log_{3}(2^{r-1})=C+(r-1)(\frac{\log 2}{\log 3}) < r-1$ when $r$ is large. The end.

Here is another solution. Lemma: If $d(an)=d(bn)$ for all $n$ then $a=b$. Proof: If $\nu_p(a)>\nu_p(b)$ then we pick $n=p^{r-1-\nu_p(a)}$ for a large prime $r$. We have $r\mid d(an)$ but $r\nmid d(bn)$ because $d(bn)=\prod\limits_{q\mid bn} \nu_q(bn)+1$. When $q\ne p$, $\nu_q(bn)+1<r$. When $q=p$, $\nu_q(bn)+1<\nu_q(an)+1=r$ as well. Claim: $d(an)=d(bn)$ for all integers $n$. Proof: Let $p\nmid abn$ be a prime. Note $d(d(anp^k))=d(d(bnp^k))$ for all $n,k$. Since $\gcd(an,p)=\gcd(bn,p)=1$, it follows that $d(d(an)(k+1))=d(d(bn)(k+1))$. Let $x=d(an), y=d(bn)$. We have $d(nx)=d(ny)$ for all $n\in \mathbb{N}$. Therefore, $d(an)=d(bn)$, as needed.

hakN wrote: Then we have $\forall k\in \mathbb{N}$, $\tau((\alpha + k)(\alpha_{1} + 1)...(\alpha_{m} + 1)) = \tau((\beta + k)(\beta_{1} + 1)...(\beta_{l} + 1))$. Now let $P$ be a large prime. Take $k$ such that $\alpha + k = 2^{P-1}$. Then, since $P$ is large, $\beta + k$ has to have a prime divisor $q$ such that $v_q(\beta + k) \equiv -1 \pmod{P}$. I don't get this part. I think it would be true if $\alpha+k$ and $(\alpha_1+1)\cdots(\alpha_m+1)$ are coprimes and $\beta+k$ and $(\beta_1+1)\cdots(\beta_l+1)$ are coprimes, but otherwise we're unsure if $P$ divides the left-hand side.