Let $a_0:=6$ and $a_{n+1} = (a_n-1)\cdot a_n\cdot (a_n+1)$. Then $a_1 = 210 = 2 \times 3 \times 5 \times 7$. What is the smallest value of $n$ for which $a_n$ is not square-free?
The congruent numbers are precisely the numbers that can be expressed as the square-free part of $x^3 - x$ where $x$ is a nonnnegative integer.
The sequence given contains some huge square-free numbers for low values of $n$. I wonder how long that pattern continues.
The question is equivalent to asking whether $b(n):=\frac{a(n)}{a(n-1)}$ are squarefree. The sequence $b(n)$ is listed in OEIS A231831 and satisfies the recurrence $b(n+1) = b(n)^3 + b(n)^2 - 1$ with $b(1)=3$. Its terms are pairwise coprime, implying that each prime divides at most one term.
Quite similarly to my treatment of Sylvester sequence, I have computed all primes $p$ below $10^{10}$ such that $p\mid b(n)$ for some $n$ (there are $16944$ such primes), and verified that $p^2\nmid b(n)$. The primes $p$ and corresponding indices $n$ are now listed in OEIS A362250 and OEIS A362251, respectively.
The question can also be analyzed heuristically by considering the map $x\mapsto x^3 + x^2 - 1$ modulo prime $p$ as random. Various kinds of statistics for such mappings and their functional graphs are given by Flajolet and Odlyzko (1989). My very rough analysis suggests that the "probability" for prime $p$ to divide some $b(n)$ is proportional to $\frac1{p^{1/2}}$, and the "probability" for $p^2$ to divide some $b(n)$ is proportional to $\frac1{p^{3/2}}$. The latter means that one should expect only a finite number of primes $p$ with $p^2\mid b(n)$, if any exists at all.
There is a somewhat similar sequence OEIS A231830 satisfying $c(n+1) = c(n)^3 - c(n)^2 + 1$ with $c(1)=5$. The above analysis applies to this sequence as well. UPDATE. I've confirmed that neither prime below $10^{10}$ (listed in OEIS A362252) divides its terms when squared.
It is interesting to consider the dynamics of the map $f\colon x\mapsto x^3-x$ on the set $X=\mathbb{Z}/p^2$. Let $C$ be the set of periodic points, i.e. those $x$ for which $f^n(x)=x$ for some $n>0$. We can divide these into orbits $C_1,\dotsc,C_m$, starting with $C_1=\{0\}$. We can then put $$ D_i=\{x\in X\setminus C: f^n(x)\in C_i\text{ for some } n > 0\}. $$ We find that $X=\coprod_{i=1}^mC_i\amalg\coprod_{i=1}^mD_i$. The problem is to prove that $6\not\in D_1$. If we put $Y=p\mathbb{Z}/p^2$ we find that for $y\in Y$ we have $f(y)=-y$ and thus $f^2(y)=y$. This shows that $Y\subseteq C$. Apart from this, experiment (with primes up to about 200) does not reveal any obvious regularities. Sometimes there are many periodic orbits in $X\setminus Y$, sometimes there are only a few, and there does not seem to be any simple formula for the size of $D_1$.
