# Equations involving particular values of the Dedekind psi function and powers of the kernel function

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 . 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  and  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 (). 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 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 .

## References:

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

 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.

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

• I can to see the article  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 $k$. – user759001 May 27 at 15:48
• 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 $5$. – Peter May 28 at 17:52
• Other solutions are $$51114852$$ and $$26622318750$$ and $$230750426250$$ – Peter May 28 at 18:11
• Hope the below answer helps, note my edit from $q^5$ to $q^6$ – Peter May 28 at 19:03
• 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. – Peter May 29 at 7:43

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

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}

• @Peter : For example, $$n=2^13^1{17}^3{1733}^5=460777037387622037854$$is a solution. It seems that you don't have this number. – mathlove May 29 at 7:35
• 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 $1$ – Peter May 29 at 7:37
• @Peter : Yes, I think the number of the solutions is $16$. I've just summarized the solutions. See the top of the answer. – mathlove May 29 at 7:49
• @user759001 : I've just added a proof of another claim about an upper bound for $N_k$. – mathlove May 29 at 14:05
• 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. – user759001 May 29 at 16:02

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 $$N=\frac{Q}{P}$$

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