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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.

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    $\begingroup$ Try to see what happes if $\frac12(3m-1)$ is very high power of some prime number $\endgroup$ – Mastrem Jun 10 '18 at 19:25
  • $\begingroup$ Many thanks for your successful hint @Mastrem I did not know the structure of the pentagonal numbers. $\endgroup$ – user243301 Jun 10 '18 at 21:22
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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

$$m_r = \frac{x_{2r+1} + 1}{6}\,.$$

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  • $\begingroup$ Many thanks, for share your nice calculations with us. I am going to study your answer tomorrow in the morning. $\endgroup$ – user243301 Jun 10 '18 at 21:19

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