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In this post we denote the Dedekind psi function as $\psi(m)$ for integers $m\geq 1$. As reference I add the Wikipedia Dedekind psi function, and [1]. One has the definition $\psi(1)=1$, and that the Dedekind psi function can be represented for a positive integer $m>1$ as $$\psi(m)=m\prod_{\substack{p\mid m\\p\text{ prime}}}\left(1+\frac{1}{p}\right).$$ On the other hand we denote the product of distinct primes dividing an integer $m>1$ as $\operatorname{rad}(m)$ see the Wikipedia Radical of an integer thus $$\operatorname{rad}(m)=\prod_{\substack{p\mid m\\p\text{ prime}}}p$$ that takes the value $1$ for $m=1$. Both functions are multiplicative.

I was inspired in [1] and [2] to state the following questions.

Question 1. I would like to know what work can be done about if the following equation (that is just an example of equation, look at next question) $$\psi(n)=\operatorname{rad}(n)^4\tag{1}$$ have finitely many solutions, when $n$ runs over positive integers greater or equal than $1$. Many thanks.

As clarification in previous RHS the expression is a fourth power. As example $$\psi(648)=\psi(2^3\cdot 3^4)=2^3\cdot\frac{3}{2}\cdot3^4\cdot\frac{4}{3}=6^4,$$ and the solutions that I know are listed here $1,648,337500,8696754$.

I don't know how to approach the problem that I think that it is similar than a problem that was in the literature ([1]). I emphasize that I'm asking for what work or heuristics can be done to know if previous equation admits finitely many solutions (after some helpful answer I should to accept the answer).

Question 2. Let $k\geq 2$ be an integer, and for each fixed $k$ we consider the solutions $n\geq 1$ of the equation $$\psi(n)=(\operatorname{rad}(n))^k\tag{2}$$ (I've added the brackets just as a redundacy). Let $$N_k=\#\{n\geq 1:n\text{ solves }\psi(n)=\operatorname{rad}(n)^k\}.$$ I would like to know if it is possible to estimate roughly the size of $N_k$ in terms of $k$ or to get, roughly, some inequality involving some bound as below $$\text{a bound in terms of }k<N_k<\text{a bound in terms of }k.$$ Many thanks.

I'm not asking for a professional statement for the estimation of $N_k$, for $k\geq 2$, just some idea about the size of $N_k$ or some inequality deduced from some mathemaical reasonings or heuristics. For this question I was inspired in Theorem from [2].

References:

[1] J. M. De Koninck, Proposed problem 10966, American Mathematical Monthly, 109 (2002), p. 759.

[2] J. M. De Koninck, F. Luca, and A. Sankaranarayanan, Positive Integers Whose Euler Function Is a Power of Their Kernel Function, Rocky Mountain J. Math. Vol. 36, No. 1 (2006), pp. 81-96.

[3] Tom M. Apostol, Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag (1976).

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  • $\begingroup$ I can to see the article [2] from Project EUCLID. I emphasize that I'm interested to study the size of $N_k$ with $k\geq 2$, just roughly to get an idea of what is the size of $N_k$ or the inequalities that these quantities $N_k$ satisfy, roughly in terms of $k$. $\endgroup$ – user759001 May 27 at 15:48
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    $\begingroup$ Maybe, this helps : The solutions besides $1$ and $648$ must have a prime factor larger than $3$, at least two prime factors and the largest must have exponent $5$. $\endgroup$ – Peter May 28 at 17:52
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    $\begingroup$ Other solutions are $$51114852$$ and $$26622318750$$ and $$230750426250$$ $\endgroup$ – Peter May 28 at 18:11
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    $\begingroup$ Hope the below answer helps, note my edit from $q^5$ to $q^6$ $\endgroup$ – Peter May 28 at 19:03
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    $\begingroup$ I posted a summary of the $10$ solutions I found but deleted it since I missed something (see below). The list , mathlove posted , should be complete. $\endgroup$ – Peter May 29 at 7:43
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Added : I've just added a proof of another claim about an upper bound for $N_k$ at the end.

This answer proves the following claim :

Claim : $$ N_k=\begin{cases}2 &\text{$\quad $if $\ k=2$} \\\\4&\text{$\quad$if $\ k=3$}\\\\ 16&\text{$\quad$if $\ k=4$} \end{cases}$$

The only solutions are $$\begin{align}k=2& : n=1,2^13^2 \\\\ k=3 & : n=1,2^{2}3^{3},2^13^25^{4},2^13^117^{4} \\\\ k=4 & : n=1,2^{3}3^{4},2^{2}3^{3}5^5,2^{2}3^{2}17^5,2^{2}3^{1}53^5, \\&\qquad\quad 2^13^311^{5},2^{1}3^{1}107^5,2^13^15^{5}17^{5},2^13^25^{4}29^{5}, \\&\qquad\quad 2^13^15^{4}89^{5},2^13^{2}5^{3}149^{5},2^13^{1}5^{3}449^{5},2^13^{3}5^{1}1249^{5}, \\&\qquad\quad 2^13^{1}17^{4}101^{5},2^13^{2}17^{3}577^{5},2^13^{1}17^{3}1733^{5}\end{align}$$

Proof :

$n=1$ is a solution for $(2)$.

For an odd prime $p$, the numerator of $\frac{p+1}{p}$ is even. This implies that if $n$ is odd larger than $1$, then the equation $(2)$ does not hold. So, $n$ has to be even, and then $n$ has a prime factor $3$.

If $n=2^s3^t$ where $s,t\ge 1$, then $(2)\implies 2^{s+1}3^t=2^k3^k\implies n=2^{k-1}3^{k}$.

If $n=2^s3^t\prod_{j=1}^{d}p_j^{e_j}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are odd primes larger than $3$, and $d,s,t,e_j$ are positive integers. Then, $(2)$ is equivalent to

$$2^s3^t\bigg(\prod_{j=1}^{d}p_j^{e_j}\bigg)\cdot\frac 32\cdot\frac 43\prod_{j=1}^{d}\bigg(1+\frac{1}{p_j}\bigg)=2^k3^k\prod_{j=1}^{d}p_j^k$$ which can be written as $$\prod_{j=1}^{d}(p_j+1)=2^{k-1-s}3^{k-t}\prod_{j=1}^{d}p_j^{k+1-e_j}$$ where we have to have $s\le k-1, t\le k$ and $e_j\le k+1$.

Since LHS is divisible at least by $2^d$, we have to have $1\le d\le k-1-s\le k-2$ implying $k\ge 3$.


$k=2$ :

The only solutions are $n=1,2^13^2$, and so $N_2=2$.


$k=3$ :

$n=1,2^{2}3^{3}$ are solutions.

If $n=2^s3^t\prod_{j=1}^{d}p_j^{e_j}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are odd primes larger than $3$, and $d,s,t,e_j$ are positive integers. Then, the equation is equivalent to

$$\prod_{j=1}^{d}(p_j+1)=2^{2-s}3^{3-t}\prod_{j=1}^{d}p_j^{4-e_j}$$ where we have to have $s\le 2, t\le 3$ and $e_j\le 4$.

Since LHS is divisible at least by $2^d$, we have to have $1\le d\le 2-s\le 1$ implying $d=1$ for which we have

$$p_1+1=2^{2-s}3^{3-t}p_1^{4-e_1}$$ Since $4-e_1=0$, we get $p_1=2^{2-s}3^{3-t}-1$ with $s=1$.

  • $2^{1}3^{1}-1=5$ is a prime, and $n=2^13^25^{4}$.

  • $2^{1}3^{2}-1=17$ is a prime, and $n=2^13^117^{4}$.

Therefore, it follows that $N_3=4$.


$k=4$ :

$n=1,2^{3}3^{4}$ are solutions.

If $n=2^s3^t\prod_{j=1}^{d}p_j^{e_j}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are odd primes larger than $3$, and $d,s,t,e_j$ are positive integers. Then, the equation is equivalent to

$$\prod_{j=1}^{d}(p_j+1)=2^{3-s}3^{4-t}\prod_{j=1}^{d}p_j^{5-e_j}$$ where we have to have $s\le 3, t\le 4$ and $e_j\le 5$.

Since LHS is divisible at least by $2^d$, we have to have $1\le d\le 3-s\le 2$.

Case 1 : $d=1$

$$p_1+1=2^{3-s}3^{4-t}p_1^{5-e_1}$$

Since $5-e_1=0$, we have $p_1=2^{3-s}3^{4-t}-1$.

  • $2^{1}3^{1}-1=5$ is a prime, and $n=2^{2}3^{3}5^5$.

  • $2^{1}3^{2}-1=17$ is a prime, and $n=2^{2}3^{2}17^5$.

  • $2^{1}3^{3}-1=53$ is a prime, and $n=2^{2}3^{1}53^5$.

  • $2^{1}3^{4}-1=161$ is not a prime.

  • $2^{2}3^{1}-1=11$ is a prime, and $n=2^13^311^{5}$.

  • $2^{2}3^{2}-1=35$ is not a prime.

  • $2^{2}3^{3}-1=107$ is a prime, and $n=2^{1}3^{1}107^5$

  • $2^{2}3^{4}-1=323$ is not a prime.

Case 2 : $d=2$

Since $s=1$, we have $$(p_1+1)(p_2+1)=2^{2}3^{4-t}p_1^{5-e_1}p_2^{5-e_2}$$ Now, $5-e_2=0$, and there is a non-negative integer $a$ such that $$p_1+1=2^13^{a}\qquad\text{and}\qquad p_2+1=2^13^{4-t-a}p_1^{5-e_1}$$

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{0}-1=17$ is a prime, and $n=2^13^15^{5}17^{5}$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{0}5^{1}-1=9$ is not a prime.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{1}5^{1}-1=29$ is a prime, and $n=2^13^25^{4}29^{5}$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{1}-1=89$ is a prime, and $n=2^13^15^{4}89^{5}$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^13^{0}5^{2}-1=49$ is not a prime.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{1}5^{2}-1=149$ is a prime, and $n=2^13^{2}5^{3}149^{5}$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{2}-1=449$ is a prime, and $n=2^13^{1}5^{3}449^{5}$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{0}5^{3}-1$ is not a prime with $3\mid p_2$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{1}5^{3}-1$ is not a prime with $7\mid p_2$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{3}-1$ is not a prime with $13\mid p_2$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{0}5^{4}-1=1249$ is a prime, and $n=2^13^{3}5^{1}1249^{5}$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{1}5^{4}-1$ is not a prime with $23\mid p_2$.

  • $p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{4}-1$ is not a prime with $7\mid p_2$.

  • $p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{0}17^{1}-1=33$ is not a prime.

  • $p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{1}17^{1}-1=101$ is a prime, and $n=2^13^{1}17^{4}101^{5}$.

  • $p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{0}17^{2}-1=577$ is a prime, and $n=2^13^{2}17^{3}577^{5}$.

  • $p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{1}17^{2}-1=1733$ is a prime, and $n=2^13^{1}17^{3}1733^{5}$.

  • $p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{0}17^{3}-1$ is not a prime with $5\mid p_2$.

  • $p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{1}17^{3}-1$ is not a prime with $7\mid p_2$.

  • $p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{0}17^{4}-1$ is not a prime with $7\mid p_2$.

  • $p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{1}17^{4}-1$ is not a prime with $5\mid p_2$.

  • $p_1=2^13^{3}-1=53$ is a prime and $p_2=2^1 3^{0}53^{1}-1$ is not a prime with $5\mid p_2$.

  • $p_1=2^13^{3}-1=53$ is a prime and $p_2=2^1 3^{0}53^{2}-1$ is not a prime with $41\mid p_2$.

  • $p_1=2^13^{3}-1=53$ is a prime and $p_2=2^1 3^{0}53^{3}-1$ is not a prime with $3\mid p_2$.

  • $p_1=2^13^{3}-1=53$ is a prime and $p_2=2^13^{0}53^{4}-1$ is not a prime with $7\mid p_2$.

Therefore, it follows that $N_4=16$.


I'm going to prove the following claim about an upper bound for $N_k$.

Claim 2 : For $k\ge 5$, $$N_k\le 2+\sum_{d=1}^{k-2}(k-2)^d(k-1)k^{d+1}(k+1)^{\frac{d(d+1)}{2}}$$

Proof :

We already know that $n=1,n=2^{k-1}3^{k}$ are solutions.

If $n=2^s3^t\prod_{j=1}^{d}p_j^{e_j}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are odd primes larger than $3$, and $d,s,t,e_j$ are positive integers. Then, $(2)$ is equivalent to

$$\prod_{j=1}^{d}(p_j+1)=2^{k-1-s}3^{k-t}\prod_{j=1}^{d}p_j^{k+1-e_j}$$ where we have to have $s\le k-1, t\le k$ and $e_j\le k+1$.

Since LHS is divisible at least by $2^d$, we have to have $1\le d\le k-1-s\le k-2$ implying $k\ge 3$.

We can write $$\begin{cases}p_1+1&=2^{a_1}3^{b_1} \\\\ p_2+1&=2^{a_2}3^{b_2}p_1^{c(2,1)} \\\\ p_3+1&=2^{a_3}3^{b_3}p_1^{c(3,1)}p_2^{c(3,2)} \\\\\qquad\vdots \\\\p_d+1&=2^{a_d}3^{b_d}p_1^{c(d,1)}p_2^{c(d,2)}\cdots p_{d-1}^{c(d,d-1)}\end{cases}$$ where $1\le a_j\le k-2,0\le b_j\le k-1$ and $0\le c(j,i)\le k$.

The number of possible $p_1$ is at most $(k-2)k$.

For each $p_1$, the number of possible $p_2$ is at most $(k-2)k(k+1)$.

For each pair $(p_1,p_2)$, the number of possible $p_3$ is at most $(k-2)k(k+1)^2$.

So, we see that the number of possible $(p_1,p_2,\cdots,p_d)$ is at most $$\prod_{j=1}^{d}(k-2)k(k+1)^{j-1}$$

For each $(p_1,p_2,\cdots,p_d)$, the number of possible $n$ is at most $$(k-1)k(k+1)^d$$

Therefore, we get, for $k\ge 5$, $$\begin{align}N_k&\le 2+\sum_{d=1}^{k-2}(k-1)k(k+1)^d\prod_{j=1}^{d}(k-2)k(k+1)^{j-1} \\\\&=2+\sum_{d=1}^{k-2}(k-2)^d(k-1)k^{d+1}(k+1)^{\frac{d(d+1)}{2}}\end{align}$$

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    $\begingroup$ @Peter : For example, $$n=2^13^1{17}^3{1733}^5=460777037387622037854$$is a solution. It seems that you don't have this number. $\endgroup$ – mathlove May 29 at 7:35
  • $\begingroup$ Yes, I saw what I missed. Maybe, you can summarize the solutions and their factorizations. We only have those $16$ solutions, right ? What I missed is, that the smaller prime factor of the form $4k+1$ need not have exponent $1$ $\endgroup$ – Peter May 29 at 7:37
  • $\begingroup$ @Peter : Yes, I think the number of the solutions is $16$. I've just summarized the solutions. See the top of the answer. $\endgroup$ – mathlove May 29 at 7:49
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    $\begingroup$ @user759001 : I've just added a proof of another claim about an upper bound for $N_k$. $\endgroup$ – mathlove May 29 at 14:05
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    $\begingroup$ I think that in this ocassion, while I study it, the fair thing is accept your answer since it solves the questions (what work can be done for Question 1 and Question 2). I'm saying it since my thoughts were that these questions should be very difficult but you and @Peter added excellent answers: the fair thing is accept your answer, many thanks again to both for share these results. $\endgroup$ – user759001 May 29 at 16:02
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Not an immediate method to estimate how many solutions below some $x$ are, but a useful method to classify the possible solutions allowing to determine all solutions upto a very large limit.

Let $S$ be a set of distinct prime numbers. Define $$P:=\prod_{p\in S} (p+1)$$

There is at most one solution $N(S)$ such that the prime factors of $N$ exactly form the set $S$. There is one solution if and only if $P$ has only prime factors belonging to $S$ and there is no prime factor $q$ with $q^6\mid P$. In this case, if we define $$Q:=\prod_{p\in S} p^5$$ the solution is $$N=\frac{Q}{P}$$

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  • $\begingroup$ Many thanks, perfect. $\endgroup$ – user759001 May 28 at 19:37
  • $\begingroup$ Many thanks again to you and the other user that added such statements. Both, incredible results! $\endgroup$ – user759001 May 29 at 16:06

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