Does anyone have a solution?

Lemma: Let $a,b,c,d$ are naturals. If there exists a non-negative integer $n$ such that $an+b,cn+d$ are both squarefree then there exists infinitely many such $n$. Proof: Let $n_0$ be the assumed solution. Then by taking $n \equiv n_0 \pmod {p^2}$, one is able to find $r$ such that for all $n = m \prod p_i^2 + r$ where $m$ varies over the naturals, $p_i^2$ does not divide $an+b$ or $cn+d$. Now let $P$ be the product of all primes up to a constant $C$ which we will fix later. Take $n = mP + r$ where $n_0 \equiv r \pmod {P^2}$. Fix a large natural $N$. We shall count the number of $m$'s where $1 \leq m \leq N$ such that one of $am+b$ or $cm+d$ is not squarefree. Clearly it suffices to consider primes $q$ where $C \leq q \leq \sqrt{max(a(mP+r)+b,c(mP+r)+d)}$. For each such $q$, if it does not divide $abcd$ then theres a total of two solutions $\pmod {q^2}$ where either $am+b$ or $cm+d$ is divisible by $q^2$ as each of them contributes one. Now wlog $c < a$. Take $C$ sufficiently large such that $C > abcd$ and also $c(NP+r)+d < a(NP+r)+b$ then an upper bound for the number of $m$'s that do not work is $\sum_{p \geq C}^{\sqrt{a(NP+r)+b}} 2 \lceil \frac{N}{p^2} \rceil < \sum_{p \geq C}^{\sqrt{a(NP+r)+b}} 2(\frac{N}{p^2} +1) = 2\pi(\sqrt{a(NP+r)+b}) + 2\sum_{p \geq C} \frac{N}{p^2}$ where $\pi(x)$ is the prime number counting function. Now if $C$ is large enough then we have $2 \sum_{p \geq C} \frac{1}{p^2} < \frac{1}{2}$ and since $\pi(x) < x$ we have that the number of $m$ that do not work is at most $O(\sqrt{N}) + \frac{N}{2}$ so the number of $m$ that works actually has a positive density and hence there are infinitely many of them. Now back to the main problem. Consider $6591n + 482$ and $1053n + 77$. Since $n = 0$ gives a solution where both are squarefree, there exists infinitely many such $n$. Let $m$ be such a solution. It is clear that $13$ and $3$ does not divide $6591n + 482$ or $1053n + 77$. Now note that $27(6591m + 482) = 169(1053m + 77) + 1$. Let $x = 27(6591m + 482)$. Then $f(x) = f(27) = 27$ and $f(x-1) = f(169(1053m + 77)) = f(169) = 26$ and hence $f(x) = f(x-1) + 1$ and we are done.

Key observation. Let there be an integer $n$, and a squarefree integer $m$ which is coprime with $n$. Then $f(nm)=f(n)$. Now say we have $f(a)=f(b)+1$ for coprime $a,b$. We can set some squarefree integers $x,y$ such that $ax=by+1$ and $(a,x)=(b,y)=1$, then we will have $$f(ax)=f(a)=f(b)+1=f(by)+1=f(ax-1)+1$$Find $a,b,x,y$ that works. Finding this is not hard. $a=27$, $b=169$, $x=482$, $y=77$ works. Now note that $x,y$ is squarefree. It now suffices to prove the following. Claim. Let there be positive integers $a,b$ and squarefree integers $x,y$. There exists infinitely many $n$ such that $an+x$ and $bn+y$ are squarefree. If this claim is true, there exists infinitely many $n$, with $ab^2n+x$ and $a^2bn+y$ squarefree. For those $n$, we will have $$f(a^2b^2n+ax)=f(a)=f(b)+1=f(a^2b^2n+by)+1=f(a^2b^2n+ax-1)+1$$which gives that $a^2b^2n+ax$ works. Now we prove the claim. Pretty much the same as above with $n_0=0$. Set $P$ as the product of all primes up to constant $C$ we will choose later. Take $n=mP$, and let $N$ be a big natural number. We count the number of $m$, $1\le m \le N$ such that at least one of $an+x$ and $bn+y$ is not squarefree. Clearly primes below $C$ fails, as for all primes $p<C$, $an+x=am^2P^2+x \equiv x \not\equiv 0 \pmod{p^2}$. Similarly, $bn+y =bm^2P^2 + y \equiv y \not\equiv 0 \pmod{p^2}$ WLOG $a>b$. Now $aNP+x>bNP+y$. We can now look at primes no larger than $\sqrt{aNP+x}$. Taking $C$ large enough, we can assure that there exists a solution to $an+x \equiv 0 \pmod{p^2}$ for $p>C$. This is similar for $bn+y$ as well. The rest is similar with this China TST problem. Below is exactly the same as above. For each prime $C \le p \le \sqrt{aNP+x}$, there are exactly two solutions to (one solution each) $amP+x \equiv 0 \pmod{q^2}$ or $bmP+y \equiv 0 \pmod{q^2}$. This implies that each prime accounts for $2\lceil \frac{N}{p^2} \rceil$, so we can work similarly with the above link. The number of $m$ that do not work is at most $$ \sum_{p \ge C}^{\sqrt{aNP+x}} 2\lceil \frac{N}{p^2} \rceil < \sum_{p \ge C}^{\sqrt{aNP+x}} 2\left(\frac{N}{p^2}+1\right) = \sum_{p\ge C} \frac{2N}{p^2} + \mathcal{O}(\sqrt{N}) < \sum_{p \ge C}^{\infty} \frac{2N}{p(p-1)} + \mathcal{O}(\sqrt{N}) < \frac{2N}{C-1} + \mathcal{O}(\sqrt {N}) < \frac{2}{3}N $$for a large enough $C$ and $N$, which gives that the number of $m$ which does work is larger than $cN$ for some constant $c$. This is enough to claim that there exists infinite $n$. We are done. $\blacksquare$

Consider the following observations. $f(27)=f(169)+1$ and $f(xy)=f(x)$ where $y$ is squarefree and $\gcd(x,y)=1$. Thus, $f(27a)=f(169b)+1$ for appropriate squarefree $a$ and $b$. Notice that $27(169m+144)=169(27m+23)+1$. Thus, if $169m+144$ and $27m+23$ are both squarefree infinitely often, we are done. Indeed, we claim $169m+144$ and $27m+23$ are both squarefree infinitely often. Indeed, recall that squarefree integers have "density/occur at a rate of" $\frac{6}{\pi^2}>\frac{1}{2}$ over all appropriate residues in any modulus. Thus, squarefree residues have "density :/" bigger than $1/2$ in both $169m+144$ and $27m+23$ implying that they coincide an infinite number of times by infinite pigeonhole.

Note that $f$ is multiplicative and $f(\text{squarefree})=1$. Further, we can easily see that $f(27)=27$ and $f(169)=26$. Consider solutions $(a,b)$ to $27a-169b=1$, which by Bezout's are given by $(a,b)=(169k+c_1,27k+c_2)$ for some positive integers $c_1,c_2$. Obviously $3 \nmid c_2$ and $13 \nmid c_1$. The idea is now to consider the number, or density of squarefree members of each arithmetic sequences, and replicate the well-known proof about the density of squarefree numbers in $\mathbb{N}$. Because it's well-known, I won't be too rigorous when adapting to arithmetic sequences. Let $\mathbb{P}$ denote the set of primes. Because $13^2$ never divides $169k+c_1$, the density of squarefree elements of $169k+c_1$ is $$\prod_{\substack{p \in \mathbb{P}\\p \neq 13}} \left(1-\frac{1}{p^2}\right)=\frac{169}{168}\prod_{p \in \mathbb{P}} \left(1-\frac{1}{p^2}\right)=\frac{169}{168} \cdot \frac{1}{\zeta(2)}>\frac{6}{\pi^2}>\frac{1}{2},$$and likewise the density of squarefree elements of $27k+c_2$ is $\frac{27}{26}\cdot \frac{1}{\zeta(2)}>\frac{1}{2}$. On the other hand, if every sufficiently large $k$ has at most one of $169k+c_1$ and $27k+c_2$ squarefree, the sum of the densities of squarefree elements is at most $1$: contradiction. It follows that infinitely many $k$ have both $a=169k+c_1$ and $b=27k+c_2$ squarefree, in which case $f(27a)=27=f(169b)+1$ and $27a=169b+1$, as desired. $\blacksquare$

Apparently you just fix $f(n)$...so much for trying to set $n = p^k$ and do LTE stuff. Notice that if $b$ is a squarefree number relatively prime to $n$, then $f(nb) = f(b)$. Furthermore, note that $f(13^2) = 26$ and $f(3^3) = 27$, so we can use this to construct an infinite set of $n$ such that $f(n) = 27$ and $f(n - 1) = 26$, hence satisfying the problem statement. To see this, I make the following claim: Claim. For any $a, b, c, d$ positive integers, if there exists $k$ such that $ak+b$ and $ck+d$ are both squarefree, then there exist infinitely many. Proof. Density. For any prime $p$, there exists some unique $p_0$ such that $k \equiv p_0 \pmod {p^2}$ yields $ak + b$ divisible by $p^2$. (Note that the condition that there exists at least one such $k$ plays into this.) Thus, for some large integer $M$, the number of $k < M$ that satisfy this is bounded by $\left \lceil \frac k{p^2} \right \rceil$, and summing this across all primes, there are at most $$\sum_{p_0 < \sqrt M} \left \lceil \frac M{p^2} \right \rceil < M\left(\frac{\pi^2}6 - 1 - \frac 14\right) < 0.5M$$such non-squarefree integers. Similarly, there are $<0.5M$ non-squarefree integers for $ck + d$, so the result follows. $\blacksquare$ Thus picking $(a, b, c, d) = (169, 144, 27, 23)$ and setting $n = 27(169k+144)$ works for each of the above $k$.

Wow this was a hilarious problem Firstly note that $f$ is multiplicative and $f(\text{squarefree})=1$. Also notice that $f(169)=26$ and $f(27)=28$. I will use this to construct infinitely. many $n$ such that $f(n-1)=26$ and $f(n)=27$. Firstly, I make the following claim Lemma(density). Given four positive integers, $a$, $b$, $c$ and $d$, if there exists an integer $k$ such that $ak+b$ and $ck+d$ are both squarefree, then there exist infinitely many. Proof. We have that for any prime $p$, there exists some unique $p_0$ such that $k \equiv p_0 \pmod {p^2}$ $\implies p^2\mid ak + b$. Hence we have that for some large integer $M$, the number of $k < M$ that satisfy this is bounded by $\left \lceil \frac {M}{p^2} \right \rceil$, and summing this across all primes, there are at most$$\sum_{p_0 < \sqrt M} \left \lceil \frac M{p^2} \right \rceil < M\left(\frac{\pi^2}6 - 1 - \frac 14\right) < 0.5M$$such non-squarefree integers. Similarly, there are $<0.5M$ non-squarefree integers for $ck + d$, so the result follows. $\blacksquare$ Now to finish, pick $(a, b, c, d) = (169, 144, 27, 23)$ and set $n=27(169k+144)$ where $k$ is described as in the above lemma

Remark