Inequalities for $\operatorname{rad}(n\text{th pentagonal number})$ and the diagonal of the regular pentagon that represents this polygonal number We denote for integers $n\geq 1$ the square-free kernel, or radical of the natural  $n>1$, as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing our integer $n>1$ with the definition $\operatorname{rad}(1)=1$.
Motivation studying the case of the squares. Then it is obvious that $$\operatorname{rad}(n^2)=\operatorname{rad}(n)\leq n,\tag{1}$$
and from this inequality we deduce a worse inequality $$\operatorname{rad}(n^2)<\sqrt{2}(n-1)\tag{2}$$
that holds for integers $n>3$. This inequality $(2)$ tell us (the obvious claim) that the square-free kernel of the perfect square $n^2$ is strictly lesser than the diagonal of the corresponding polygonal number, the square number $$\text{Square Number}(n)=n^2.$$
In this post we explore the similar inequality than $(2)$ for pentagonal numbers: seems that the most of the naturals $n> 1$ satisfy the inequality
$$\operatorname{rad}\left(\frac{3n^2-n}{2}\right)>\text{diagonal of regular pentagon that represents the }n\text{th pentagonal number},$$
that is
$$\operatorname{rad}\left(\frac{3n^2-n}{2}\right)>\left(\frac{1+\sqrt{5}}{2}\right)(n-1).\tag{3}$$

Question. But also seems that there exist many integers $m>1$ satisfying $$\operatorname{rad}\left(\frac{3m^2-m}{2}\right)<\left(\frac{1+\sqrt{5}}{2}\right)(m-1).\tag{4}$$
Prove or refute that there exist infinitely many integers $m>1$ for which the inequality $(4)$ holds.
Many thanks.

I hope that there aren't mistakes in my presentation of previous problem and that is well motivated. I've calculated with the help of a Pari/GP program the first few terms of the sequence corresponding to $(4)$, but I don't see an easy pattern  to hypothesize a family of infinitely many solutions of $(4)$.
As references I add here, for example, the Wikipedia's article dedicated to Pentagonal numbers and the MathWorld's article dedicated to Pentagon.
 A: If the $m^{\text{th}}$ pentagonal number is a square, then we have $\DeclareMathOperator{\rad}{rad}$
$$\rad \biggl(\frac{3m^2-m}{2}\biggr) \leqslant \sqrt{\frac{3m^2-m}{2}} \leqslant \sqrt{\frac{3}{2}}\cdot m\,.$$
Since $\sqrt{\frac{3}{2}} < \frac{3}{2} < \frac{1 + \sqrt{5}}{2}$, in this case we will have $(4)$ except for a few very small $m$. So if there are infinitely many pentagonal numbers that are also squares, $(4)$ holds for infinitely many $m$.
Hence let's look when a pentagonal number is a square.
\begin{align}
&& \frac{3m^2-m}{2} &= k^2 \\
&\iff& 3m^2 - m &= 2k^2 \\
&\iff& 36m^2 - 12m &= 24k^2 \\
&\iff& (6m - 1)^2 &= 24k^2 + 1 \\
&\iff& (6m-1)^2 - 24k^2 &= 1
\end{align}
One knows that Pell's equation $x^2 - Dy^2 = 1$ always has infinitely many solutions (where $x$ and $y$ are positive integers) if $D$ is not a perfect square. Since $24$ is not a perfect square, it remains to see that of the infinitely many solutions of $x^2 - 24 y^2 = 1$, infinitely many have $x \equiv 5 \pmod{6}$. The solutions are given by
$$x_r + y_r\sqrt{24} = (5 + \sqrt{24})^r\,,$$
so we have the recurrence
$$x_{r+1} + y_{r+1}\sqrt{24} = (x_r + y_r\sqrt{24}) (5 + \sqrt{24}) = (5x_r + 24y_r) + (x_r + 5y_r)\sqrt{24}$$
and in particular
$$x_{r+1} \equiv 5x_r \pmod{6}\,.$$
Since $x_1 = 5 \equiv 5 \pmod{6}$, the solutions
$$x_{2r+1} + y_{2r+1}\sqrt{24} = (5 + \sqrt{24})^{2r+1}$$
yield pentagonal square numbers for
$$m_r = \frac{x_{2r+1} + 1}{6}\,.$$
