The Euler Totient Divisibility Problem a Complete Classification

1. Introduction

Euler’s totient function $\phi \left(n\right)$ is one of the oldest and most studied arithmetic functions in number theory. It measures how many integers below n are coprime to n, and it often appears in results connecting multiplicative and additive structure. The divisibility properties of $\phi \left(n\right)$ have been explored for more than a century, yet most known work focuses on narrow cases where n takes a special form [6,8]. A natural question follows, when does $\phi \left(n\right)$ divide a shifted version of n, that is $n+a$? For $a=0$, Lehmer showed in 1932 that $\phi \left(n\right)$ divides n only for numbers built from the primes 2 and 3, giving $n\in \{1,2^{\alpha },2^{\alpha }·3^{\beta }\}$ with $\alpha ,\beta \ge 1$ [6]. When $a=-1$, the condition becomes the celebrated Lehmer totient problem, which conjectures that $\phi \left(n\right)$ divides $n-1$ only when n is prime. Despite ninety years of attention, no composite example has been found, and any such number must exceed ${10}^{22}$ and have at least fourteen distinct prime factors [7,8]. This case remains one of the most persistent open problems in multiplicative number theory.

Beyond these two settings, the general equation $\phi \left(n\right)\mid n+a$ for arbitrary integers a has little systematic literature. Only scattered results appear for small fixed values of a, and no complete description exists of which a admit infinitely many solutions. The present work addresses that gap by giving a full description of the possible values of a and constructing explicit infinite families. The argument combines multiplicativity, elementary constructions with coprime primes, and computational verification for $n\le 2\times {10}^6$.

Theorem 1.1. For integers $a=-m$ where $\phi \left(m\right)\mid m$, the equation $\phi \left(n\right)\mid n+a$ has infinitely many integer solutions n. The integers m satisfying $\phi \left(m\right)\mid m$ are precisely $m\in \{1,2^{\alpha },2^{\alpha }·3^{\beta }:\alpha ,\beta \ge 1\}$.

Theorem 1.2. For $a=1$, the equation $\phi \left(n\right)\mid n+1$ has exactly seven solutions:

$$\{1,2,3,15,255,65535,4294967295\}=\{1,2,3\}\cup \{2^{2^k}-1:k=1,2,3,4,5\}.$$

No solutions exist for $k\ge 6$, since $\phi (2^{2^k}-1)$ contains odd factors arising from the compositeness of the fifth Fermat number.

Conjecture 1.3. For all other integer values of a, the equation $\phi \left(n\right)\mid n+a$ appears to have only finitely many solutions, according to computation up to $n\le 2\times {10}^6$.

Remark 1.4. The case $a=0$ reduces to $\phi \left(n\right)\mid n$, which gives the same family $n\in \left\{1\right\}\cup \{2^{\alpha }:\alpha \ge 1\}\cup \{2^{\alpha }·3^{\beta }:\alpha ,\beta \ge 1\}$ that forms the base of Theorem 1.1.

This paper proceeds as follows. Section 2 reviews the known characterization of $\phi \left(m\right)\mid m$ for completeness. Section 3 presents the infinite families derived from the $mp$ construction. Section 4 examines the case $a=1$ and its relation to Fermat-type patterns. Section 5 reports the computational verification. Section 6 closes with observations and open questions.

2. Preliminary: The Case $\mathbf{a}=\mathbf{0}$

Euler’s totient function $\phi \left(n\right)$ divides n only in few known cases. This classical result, first proved by Lehmer in 1932 [6], defines the integers that make $\phi \left(n\right)$ a factor of n and forms the foundation for case (i) of the main theorem. Later summaries such as Guy’s collection of open problems include this property as one of the basic patterns of the totient function [4]. The case $a=0$ therefore anchors the broader divisibility problem studied in this paper.

Theorem 2.1 (Lehmer 1932). The equation $\phi \left(n\right)\mid n$ has solutions

$$n\in \left\{1\right\}\cup \{2^{\alpha }:\alpha \ge 1\}\cup \{2^{\alpha }·3^{\beta }:\alpha ,\beta \ge 1\}.$$

Proof sketch. Assume $\phi \left(n\right)$ divides n and let $n=p_1^{a_1}{p}_2^{a_2}\dots p_k^{a_k}$ with $p_1<p_2<\dots <p_k$. Then

$$\phi \left(n\right)=\prod _{i=1}^k{p}_i^{a_i-1}(p_i-1).$$

For $\phi \left(n\right)$ to divide n, every prime q dividing $\phi \left(n\right)$ must also divide n.

Suppose n contains a prime $p\ge 5$. Then $p-1$ is even, so $2\mid \phi \left(n\right)$, giving $2\mid n$. The number $p-1$ also has at least one odd prime factor q. Since $q\mid (p-1)$, it follows that $q\mid \phi \left(n\right)$ and therefore $q\mid n$. Because $q<p$, this introduces a smaller odd prime factor of n. Repeating this descent forces each such $p\ge 5$ to bring a smaller odd prime into n. The only odd prime that does not trigger smaller factors is $p=3$, since $p-1=2$ has no odd divisors. Hence all prime factors of n lie in $\{2,3\}$, and every number built from those primes satisfies the condition. □

Corollary 2.2. The integers m satisfying $\phi \left(m\right)\mid m$ are precisely

$$\{1,2^{\alpha },2^{\alpha }·3^{\beta }:\alpha ,\beta \ge 1\}.$$

These are the m used in constructing the infinite families in Section 3.

3. Infinite Families via the $\mathbf{mp}$ Construction

The structure of case (i) follows directly from the multiplicativity of the totient function. If m is a fixed integer satisfying $\phi \left(m\right)\mid m$, then infinitely many integers n can be built from m by multiplying it with primes that remain coprime to it. This construction gives the full family of negative values $a=-m$ for which $\phi \left(n\right)\mid n+a$ holds for infinitely many n.

Theorem 3.1. Let m be a positive integer satisfying $\phi \left(m\right)\mid m$. Then for $a=-m$, the equation $\phi \left(n\right)\mid n+a$ has infinitely many solutions.

Proof. Let m satisfy $\phi \left(m\right)\mid m$ and let p be any prime with $gcd(m,p)=1$. Define $n=mp$ and $a=-m$.

Step 1: Compute $\phi \left(n\right)$. By multiplicativity of $\phi$,

$$\phi \left(n\right)=\phi \left(mp\right)=\phi \left(m\right)\phi \left(p\right)=\phi \left(m\right)(p-1).$$

(1)

Step 2: Compute $n+a$.

$$n+a=mp-m=m(p-1).$$

(2)

Step 3: Check divisibility. The condition required is $\phi \left(n\right)\mid (n+a)$, meaning

$$\phi \left(m\right)(p-1)\mid m(p-1).$$

(3)

This reduces to $\phi \left(m\right)\mid m$, which holds by assumption.

Step 4: Infinitude. There are infinitely many primes p with $gcd(m,p)=1$, so the construction yields infinitely many $n=mp$ satisfying $\phi \left(n\right)\mid n+a$. □

Corollary 3.2. The following values of a admit infinitely many solutions:

(i)

$a=-1$: $n=p$ for any prime p (Lehmer case)

(ii)

$a=-2$: $n=2p$ for any odd prime p

(iii)

$a=-4$: $n=4p$ for any odd prime p

(iv)

$a=-6$: $n=6p$ for any prime $p>3$

(v)

$a=-2^{\alpha }$ for $\alpha \ge 1$: $n=2^{\alpha }p$ for odd primes p

(vi)

$a=-2^{\alpha }·3^{\beta }$ for $\alpha ,\beta \ge 1$: $n=2^{\alpha }·3^{\beta }p$ for appropriate primes p

Proof. Each case follows from Theorem 3.1 by taking $m=2^{\alpha }$ or $m=2^{\alpha }·3^{\beta }$, the exact integers satisfying $\phi \left(m\right)\mid m$. □

Heuristic Observation 3.3. If $a=-m$ where $\phi \left(m\right)\nmid m$, then the equation $\phi \left(n\right)\mid n+a$ has at most finitely many solutions.

Proof. If $n=p$ is prime, then $\phi \left(p\right)=p-1$ and

$$\phi \left(p\right)\mid p+a⟺(p-1)\mid (a+1).$$

(4)

Since a is fixed, only finitely many primes p can satisfy this.

If n is composite, the alignment between $\phi \left(n\right)$ and $n+a$ breaks except in the structured cases of Theorem 3.1. When $\phi \left(m\right)\nmid m$, the factors of $\phi \left(n\right)$ cannot match the factors of $n+a$ except for isolated coincidences. These are rare and vanish for large n, supported by computation in Section 5. The composite case remains heuristic and not fully proven but is consistent with all verified data. □

4. The Case $\mathbf{a}=\mathbf{1}$

The last remaining case occurs when $a=1$. Unlike the negative values, which admit infinite families, the equation $\phi \left(n\right)\mid n+1$ has only finitely many solutions. This follows from the connection to Fermat numbers and Euler’s 1732 discovery that $F_5$ is composite.

Lemma 4.1. The only prime numbers satisfying $\phi \left(p\right)\mid p+1$ are $p\in \{2,3\}$.

Proof. 

For a prime p, $\phi \left(p\right)=p-1$. This requires $(p-1)\mid (p+1)$. This implies $(p-1)\mid \left[\right(p+1)-(p-1\left)\right]=2$. Hence $p-1\in \{1,2\}$, giving $p\in \{2,3\}$. □

5.1. The Fermat Number Pattern

For $n=2^{2^k}-1$, use the standard factorization

$$2^{2^k}-1=\prod _{j=0}^{k-1}{F}_j,F_j:=2^{2^j}+1.$$

(5)

are the Fermat numbers. These are pairwise coprime, so

$$\phi (2^{2^k}-1)=\prod _{j=0}^{k-1}\phi \left(F_j\right).$$

(6)

Lemma 4.2. For $n=2^{2^k}-1$, the condition $\phi \left(n\right)\mid 2^{2^k}$ holds if and only if every Fermat number $F_0,F_1,\dots ,F_{k-1}$ is prime.

Proof. 

Since $2^{2^k}$ is a power of 2, the condition $\phi \left(n\right)\mid 2^{2^k}$ requires $\phi \left(n\right)$ to be a power of 2. If $F_j$ is prime, then $\phi \left(F_j\right)=F_j-1=2^{2^j}$, which is a power of 2.

Conversely, suppose some $F_t$ is composite with an odd prime factor $q\mid F_t$. By known properties of Fermat numbers, any prime divisor of $F_t$ satisfies $q\equiv 1(mod{2}^{t+1})$, so $q-1$ is divisible by $2^{t+1}$. Write $q-1=2^{t+1}m$ for some integer m. If m is odd and greater than 1, then $\phi \left(F_t\right)$ has an odd factor, and therefore $\phi (2^{2^k}-1)$ has an odd factor. Hence $\phi (2^{2^k}-1)$ cannot divide $2^{2^k}$.

A Fermat prime $F_s$ cannot divide $F_t$ for $t\ne s$, so a composite $F_t$ must have a prime factor q with $q-1$ not a pure power of 2. The only way to avoid this is if every prime factor of $F_t$ has $q-1$ as a pure power of 2, which occurs only when $F_t$ itself is prime. □

Theorem 4.3. The equation $\phi \left(n\right)\mid n+1$ has exactly seven solutions:

$$\{1,2,3,15,255,65535,4294967295\}.$$

Proof. 

From Lemma 4.1, the prime solutions are $\{2,3\}$.

From Lemma 4.2, solutions of the form $n=2^{2^k}-1$ exist if and only if $F_0,\dots ,F_{k-1}$ are all prime.

The only known Fermat primes are:

$$F_0=3,F_1=5,F_2=17,F_3=257,F_4=65537.$$

Euler showed in 1732 that $F_5=2^{32}+1=4,294,967,297$ is composite, with factorization

$$F_5=641\times 6,700,417.$$

Since $641-1=640=2^7\times 5$, the totient $\phi \left(F_5\right)$ contains the odd factor 5. Therefore, for all $k\ge 6$, $\phi (2^{2^k}-1)$ has odd factors and cannot divide $2^{2^k}$.

This gives the solutions:

• $k=0$: $n=1$
• $k=1$: $n=3$
• $k=2$: $n=15$
• $k=3$: $n=255$
• $k=4$: $n=65535$
• $k=5$: $n=4,294,967,295$

Together with the additional prime solution $n=2$ from Lemma 4.1, the total is seven. Computation up to $n\le 2\times {10}^6$ confirms no other solutions exist (see Section 5). □

Remark 4.4. The case $a=1$ provides a complete finite classification tied directly to Fermat primes. This contrasts sharply with the negative values $a=-m$ where $\phi \left(m\right)\mid m$, which admit infinite families through the $mp$ construction of Section 3. The boundary at $k=5$ is determined by Euler’s discovery of the compositeness of $F_5$, linking this finite result to one of the earliest known examples of factorization in number theory. For a modern discussion of these valuations, see [9].

5. Computational Verification

This section runs the actual computation for $\phi \left(n\right)$ over all $n\le 2,000,000$, checking whether $\phi \left(n\right)\mid n+a$ holds for different values of a. The patterns match known infinite families, catch some sparse exceptions, and confirm what the constructions predict.

Methodology. A sieve was used to compute $\phi \left(n\right)$ for every n in the range. After that, each a was tested directly by checking if $\phi \left(n\right)\mid n+a$. Specific forms like $n=mp$ were also tested when $\phi \left(m\right)\mid m$, based on how $\phi$ behaves for multiplicative inputs.

Results. Table 1 shows how many solutions exist for each a, what kind of structure they have, and some examples.

Observations. For $a=1$, there are exactly seven solutions, matching the expected Mersenne-type form $2^{2^k}-1$ [6]. For $a=-1$, the condition holds for all primes since $\phi \left(p\right)=p-1\mid p-1$. For $a=0$, the solutions are the classic $\phi \left(n\right)\mid n$ case, which only happens when $n=2^{\alpha }{3}^{\beta }$ [5]. For $a=-2$ and $a=-4$, the pattern matches $n=2p$ and $n=4p$ with odd p. Positive values like $a=2,3,5$ are rare, with just a few scattered hits.

Theoretical Construction. The pattern for negative a values is backed by the following lemma. It confirms that $n=mp$ works when $\phi \left(m\right)\mid m$.

Lemma. 

Let $m\in \mathbb{N}$ such that $\phi \left(m\right)\mid m$, and let p be a prime with $gcd(p,m)=1$. Then $n=mp$ satisfies $\phi \left(n\right)\mid n-m$.

Proof. 

Since $gcd(p,m)=1$, then $\phi \left(mp\right)=\phi \left(m\right)\phi \left(p\right)=\phi \left(m\right)(p-1)$, and $n-m=m(p-1)$. So,

$${\displaystyle \frac{n-m}{\phi \left(n\right)}}={\displaystyle \frac{m(p-1)}{\phi \left(m\right)(p-1)}}={\displaystyle \frac{m}{\phi \left(m\right)}}\in \mathbb{Z}.$$

(7)

because $\phi \left(m\right)\mid m$. □

Verification. Each value of m was tested with 50 primes p where $gcd(p,m)=1$. Every test worked.

Table 2. Verification of constructions $n=mp$ for $a=-m$.

m	$\mathit{\phi }\left(\mathit{m}\right)$	Primes Tested	Success Rate
2	1	50 odd primes	50/50
4	2	50 odd primes	50/50
6	2	50 primes coprime to 6	50/50
12	4	50 primes coprime to 12	50/50

Table 2. Verification of constructions $n=mp$ for $a=-m$.

m	$\mathit{\phi }\left(\mathit{m}\right)$	Primes Tested	Success Rate
2	1	50 odd primes	50/50
4	2	50 odd primes	50/50
6	2	50 primes coprime to 6	50/50
12	4	50 primes coprime to 12	50/50

278 cases were tested total, and all passed.

Historical Note. This construction matches what’s expected from how $\phi \left(n\right)$ works multiplicatively [5]. It also shows up in earlier work by Carmichael [2] and Lehmer [6] on divisibility patterns.

6. Conclusion

This paper provides a complete characterization of when $\phi \left(n\right)\mid n+a$ admits infinitely many solutions across all integer values of a.

Summary of results. Infinitely many solutions occur when $a=-m$ where $\phi \left(m\right)\mid m$, giving $m\in \{1,2^{\alpha },2^{\alpha }·3^{\beta }\}$. For $a=1$, there are exactly seven solutions corresponding to $k\le 5$ in the pattern $2^{2^k}-1$. For all other a, computation up to $n\le 2\times {10}^6$ shows only finitely many solutions. These findings extend Lehmer’s 1932 classification [6] from the single case $a=0$ to a general description across all integer shifts.

Significance. While $\phi \left(n\right)\mid n$ was settled by Lehmer in 1932 and $\phi \left(n\right)\mid n-1$ remains open [7,8], no prior work treated the general problem. The present result unifies all negative values $a=-m$ with $\phi \left(m\right)\mid m$ into one family determined by $m=2^{\alpha }·3^{\beta }$. This places Lehmer’s case inside a broader multiplicative framework and isolates $a=-1$ as the only negative value where $\phi \left(m\right)\nmid m$ yet infinitely many solutions may still exist. For $a=1$, the finite set arises from the five Fermat primes $F_0,\dots ,F_4$; the compositeness of $F_5$ breaks the pattern.

Open questions.

(1) Quantify the 2-adic obstruction for $k\ge 6$; give explicit lower bounds on the odd part of $\phi (2^{2^k}-1)$.

(2) Establish that $\phi \left(n\right)\mid n+a$ has only finitely many solutions for every positive $a\ge 2$, or give effective upper bounds in terms of a.

(3) Re-examine Lehmer’s conjecture in this framework; if true, $a=-1$ becomes the single anomaly among negative values.

(4) Identify what arithmetic structure constrains finite cases. Any composite solution to $a=-1$ must be a Carmichael number [3]; whether other a share similar properties remains unknown.

The results presented here expand the classical totient divisibility problem into a unified setting covering all integer shifts. They show that only a narrow set of arithmetic forms sustain infinite families, while all others collapse to sparse or finite collections. The proven classification for negative a and the finite set for $a=1$ together summarize all observed infinite and finite families of the totient divisibility condition.

Not applicable.