Squares in $(\operatorname{rad}(1)^2+1)\cdot(\operatorname{rad}(2)^2+1)\ldots(\operatorname{rad}(n)^2+1)$

For integers $n\geq 1$, $\operatorname{rad}(n)$ denotes the radical of an integer, see in Wikipedia this definition $$\operatorname{rad}(n)=\prod_{p\mid n}p,$$ if $n>1$ with factorization $n=\prod_{p\mid n}p^{e_p}$, and defining $\operatorname{rad}(1)=1$.

I’m inspired in a similar conjecture due to Amdeberhan, and Medina and Moll.

Question. I believe that the following conjecture holds. Can you prove or refute it?

There exists an integer $N$ such that $\forall n>N$ $$\prod_{k=1}^n(1+(\operatorname{rad}(k))^2)$$ is not a square.

Many thanks.


Amdeberhan, and Medina and Moll, Arithmetical properties of a sequence arising from an arctangent sum, J. Number Theory (2007).

Solutions Collecting From Web of "Squares in $(\operatorname{rad}(1)^2+1)\cdot(\operatorname{rad}(2)^2+1)\ldots(\operatorname{rad}(n)^2+1)$"

Important Note: The argument detailed below proves the desired theorem if the quantity $$C_{\text{rad}}=\frac{6}{\pi^2}\sum_{l=1}^\infty \frac{d_l}{l(l+1)}$$ satisfies $C_{\text{rad}}>\frac{1}{2}$, where $d_l$ is the relative density of squarefree integers $k$ in the interval $\frac{n}{l+1}<k\leq \frac{n}{l}$ with $\text{num_rad}_n(k)=\text{odd}$, where $\text{num_rad}_n$ is defined below. Currently I can prove that $$0.46684<C_{\text{rad}}<0.522,\ \ \ \ \ \ \ \ \ \ (1)$$ but I do not know how to evaluate this series exactly. If $C_{\text{rad}}<\frac{1}{2}$, then this proof technique fails to achieve the desired result.

For squarefree integers $k$, define the function
$$\text{num_rad}_{n}(k)=\#\left\{ j\leq n:\ \text{rad}(j)=\text{rad}(k)\right\}.$$
Since squarefree integers have unique radicals, it follows that
k\leq n\\
k\text{ squarefree}
Split the squarefree integers below $n$ into two subsets, $S_{n}^{\text{odd}}$ and $S_{n}^{\text{even}},$ defined by
$$S_{n}^{\text{odd}}=\left\{ k\leq n\text{ squarefree}:\ \text{num_rad}_{n}(k)\text{ is odd}\right\}$$
$$S_{n}^{\text{even}}=\left\{ k\leq n:\ \text{num_rad}_{n}(k)\text{ is even}\right\}.$$
$$\prod_{k=1}^{n}\left(1+(\text{rad}(k))^{2}\right)=K^{2}\prod_{k\in S_{n}^{\text{odd}}}\left(1+(\text{rad}(k))^{2}\right)$$
for some integer $K$, where $\text{rad}(k)\neq\text{rad}(j)$ for any distinct $j,k\in S_{n}^{\text{odd}}$. Our goal is to lower bound the size of the squarefree part of the quantity

$$Q(n)=\prod_{k\in S_{n}^{\text{odd}}}\left(1+(\text{rad}(k))^{2}\right),$$

and we do so in the following way:

  • Step 1: Upper bound the size of the squared part using an argument similar to the answer here: On prime factors of $n^2+1$
  • Step 2: Lower bound the size of $Q(n)$.

Unfortunately the lower bound may not be large enough, and relies on the constant $C_{\text{rad}}$ being greater than $1/2$.

Bounding the Squareful Part

Proposition 1: Suppose that $Q(n)=M^2\cdot R$ where $R$ is squarefree. Then $$\log M\lesssim \frac{n\log n}{2}$$

Proof: We first establish that if $p|M^2$ then $p\leq 2n$. If $p$ divides only one term $(1+\text{rad}(k)^2)$ then we must have $p^2$ dividing that term, and so $p\leq 2\text{rad}(k)\leq 2n$. If $p$ divides two different terms, $(1+\text{rad}(k)^2)$, $(1+\text{rad}(j)^2)$ then $p$ divides $(\text{rad}(k)-\text{rad}(j))(\text{rad}(j)+\text{rad}(k))$ and so once again $p\leq 2n$. It is a theorem of Fermat that if $p$|(m^2+1) for an integer $m$, then $p\equiv 1\pmod{4}$. This congruence fact is critical, as it allows us to gain a factor of $2$ in the exponent over the trivial bound.

Thus, we may write $$M^2=\prod_{\begin{array}{c}
p\equiv1\ (4)
\end{array}}p^{\alpha_{p}}$$ where each $\alpha_{p}$ is even. Now, among the integers $1,\dots,p^k$, there can be at most two for which $p^k|x^2+1$, and so for each $p\equiv 1\pmod{4}$ $$\alpha_{p}\leq\sum_{j\leq\log_{p}(n^{2}+1)}2\lceil\frac{n}{p^{j}}\rceil\leq2\log_{p}(n^{2}+1)+\frac{2n}{p-1}.$$ This yields the upper bound $$\log M^2\leq\sum_{\begin{array}{c}
p\equiv1\ (4)
\end{array}}\left(\frac{2n\log p}{p-1}+2\log(n^{2}+1)\right),$$
$$\leq 2\pi(n,4,1)\log(n^2+1)+2n\sum_{\begin{array}{c}
p\equiv1\ (4)
\end{array}}\frac{\log p}{p-1}.$$
Asymptotically, by the prime number theorem for arithmetic progressions, it follows that

$$\log M^2 \lesssim n\log n.$$

Lower Bounding $\log Q(n)$

Let $k$ be a squarefree number. Then we want to understand how many integers below $n$ have $k$ as their radical, that is the number of divisors $d|k$ such that $d\cdot k\leq n.$ Let $M_{n}^{l}=\left\{ k\text{ squarefree}:\ \frac{n}{l+1}\leq k\leq\frac{n}{l}\right\},$ where $l\geq1$ is an integer. We split into these ranges since $$d\cdot k=\begin{cases}
\leq n & \text{if }d\leq l\\
>n & \text{if }d>l
\end{cases},$$ which means that we are working a truncated divisor function. In what follows, we will attempt to calculate the density $d_{l}$ of squarefree integers $k$ in $M_{n}^{l}$ that also lie in $S_{n}^{\text{odd}}$, that is that have $\text{num\_rad}_{n}(k)$ odd. With this notation, it follows that $$\frac{|S_{n}^{\text{odd}}|}{n}\sim\frac{6}{\pi^{2}}\cdot\left(\sum_{l=1}^{\infty}\frac{d_{l}}{l(l+1)}\right)=C_{\text{rad}},$$ where the $\frac{6}{\pi^{2}}$ factor appears since it is the density of the squarefree integers.

Taking the product over $l$ of the smallest element in $M_l$ to the power of the number of elements in this interval, interval, we find that $$\log(Q(l))\geq \frac{6}{\pi^2} \sum_{l=1}^\infty \frac{d_l}{l(l+1)}\log\frac{n}{l+1}=C_{\text{rad}} n\log n+O(n),$$ and so we have proven the following proposition:

Proposition 2: We have that $$\log(Q(n))\geq C_{\text{rad}} n\log n+O(n).$$

Combining proposition $1$ and $2$ we immediately deduce:

Theorem 1: If $C_{\text{rad}}>\frac{1}{2}$, then there exists a large constant $N_0$ such that for all $n>N_0$
$$\prod_{k=1}^n \left(1+\left(\text{rad}(k)\right)^2\right)$$ is never a square.

Calculations: Calculating $d_l$ for $l=1,2,3,4$

When $l=1,$ all the integers in $M_{n}^{l}$ lie in $S_{n}^{\text{odd}}$ since $d=1$ is the only allowable divisor, and hence $$d_{1}=1.$$ When $l=2,$ then those squarefree integers divisible by $2$ must be excluded. This amounts to multiplying by $\left(1-\frac{1}{2}\right),$ but since the squarefree integers already excluded multiples of $4,$ we must divide by $\left(1-\frac{1}{2^{2}}\right),$ and so $$d_{2}=\frac{\left(1-\frac{1}{2}\right)}{\left(1-\frac{1}{2^{2}}\right)}=\frac{2}{3}.$$ When $l=3,$ we must avoid factors of $2,$ factors of $3,$ but not factors of $2$ and $3,$ since in that case we have an odd number of divisors. Thus $$d_{3}=\frac{\left(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)+\frac{1}{2\cdot3}\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\right)}{\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)}=\left(\frac{7}{12}\right).$$ When $l=4,$ we must only avoid factors of $3,$ and so $$d_{4}=\frac{\left(1-\frac{1}{3}\right)}{\left(1-\frac{1}{3^{2}}\right)}=\frac{3}{4}.$$ In general, we will have that $$d_l=\prod_{p\leq l}\left(1-\frac{1}{p}\right)\cdot \prod_{p\leq l}\left(1-\frac{1}{p^2}\right)^{-2}\cdot \left(\sum_{n}\frac{f_l(n)}{n}\right)$$ for a very specific function $f_l(n)$ taking values $0$ or $1$, and which is always $0$ if $p>l$ divides $n$ or if $n$ is not squarefree, and is $1$ if $n=1$. Cleaning this up a little, $$d_l=\prod_{p\leq l}\left(1+\frac{1}{p}\right)^{-1}\left(\sum_{n=1}^\infty\frac{f_l(n)}{n}\right).$$ In particular, for primes $q$, we have that $$f_l(q)=\sum_{q^k\leq l}(-1)^{k-1},$$ and in general it seems fairly messy.

Explicit Calculations:

  • $l=1:$ $d_l=1$
  • $l=2:$ $d_l=\frac{2}{3}$
  • $l=3:$ $d_l=\frac{7}{12}$
  • $l=4:$ $d_l=\frac{3}{4}$
  • $l=5:$ $d_l=\frac{2}{3}$ [Wolfram Link]
  • $l=6:$ $d_l=\frac{13}{18}$ [Wolfram Link]
  • $l=7:$ $d_l=\frac{2}{3}$ [Wolfram Link]
  • $l=8:$ $d_l=\frac{1}{2}$ [Wolfram Link]
  • $l=9:$ $d_l=\frac{5}{8}$ [Wolfram Link]
  • $l=10:$ $d_l=\frac{29}{48}$ [Wolfram Link]

Using these numbers, we obtain the bounds stated in equation $(1)$.