Let $n=am+1$ where $a $ and $m>1$ are positive integers and let $p$ be the least prime divisor of $m$. Prove that if $a<p$ and $ m \ | \ \phi(n)$ then $n$ is prime.

This question is a generalisation of the question at Let $n=apq+1$. Prove that if $pq \ | \ \phi(n)$ then $n$ is prime.. Here the special case when $m$ is a product of two distinct odd primes has been proven. The case when $m$ is a prime power has also been proven here https://arxiv.org/abs/2005.02327.

How do we prove that the proposition holds for an arbitrary positive integer integer $m>1 $? ( I have not found any counter - examples).

Note that if $n=am+1$ is prime, we have $\phi(n)= n-1=am$. We see that $m \ | \ \phi(n) $. Its the converse of this statement that we want to prove i.e. If $m \ | \ \phi(n) $ then $n$ is prime.

If this conjecture is true, then we have the following theorem which is a generalisation ( an extension) of Lucas's converse of Fermat's little theorem.

$\textbf {Theorem} \ \ 1.$$ \ \ \ $ Let $n=am+1$, where $a$ and $m>1$ are positive integers and let $p$ be the least prime divisor of $m$ with $a<p$. If for each prime $q_i$ dividing $m$, there exists an integer $b_i$ such that ${b_i}^{n-1}\equiv 1\ (\mathrm{mod}\ n)$ and ${b_i}^{(n-1)/q_i} \not \equiv 1(\mathrm{mod}\ n)$ then $n$ is prime.

Proof. $ \ \ \ $ We begin by noting that ${\mathrm{ord}}_nb_i\ |\ n-1$. Let $m={q_1}^{a_1}{q_2}^{a_2}\dots {q_k}^{a_k}$ be the prime power factorization of $m$. The combination of ${\mathrm{ord}}_nb_i\ |\ n-1$ and ${\mathrm{ord}}_nb_i\ \nmid (n-1)/q_i$ implies ${q_i}^{a_i}\ |\ {\mathrm{ord}}_nb_i$. $ \ \ $${\mathrm{ord}}_nb_i\ |\ \phi (n)$ therefore for each $i$, ${q_i}^{a_i}\ |\ \phi (n)$ hence $m\ |\ \phi (n)$. Assuming the above conjecture is true, we conclude that $n$ is prime.

Taking $a=1$, $m=n-1$ and $p=2$, we obtain Lucas's converse of Fermat's little theorem. Theorem 1 is thus a generalisation (an extension) of Lucas's converse of Fermat's little theorem.

On recommendation by the users, this question has been asked on the MathOverflow site, https://mathoverflow.net/questions/373497/prove-that-there-are-no-composite-integers-n-am1-such-that-m-phin

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    If it took research paper to even solve a special case, then this question might be better suited for mathoverflow – supinf Sep 30 '20 at 10:52
  • Hello! David Jones, may I please make an article about this very beautiful problem? –  Sep 30 '20 at 11:00
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    @supinf, let's not neglect the possibility that someone could come up with a beautiful short proof or find a counterexample. – ASP Sep 30 '20 at 11:44
  • @Vlad. This problem is already part of the article provided in the question link. You may consider first looking into it then expound on it on a different article. It's okay – ASP Sep 30 '20 at 11:47
  • the problem has simply been stated in the article. – ASP Sep 30 '20 at 11:47
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    I saw, I think I am close to proving it, on the general case, and would like to make my own paper. Thanks. I will notify you –  Sep 30 '20 at 12:03
  • You cannot prove that the proposition holds for an arbitrary positive integer $ m $. From the conditions it must be $ p \neq 2 $ so $ m $ is odd and $ a $ must be even. – user140242 Sep 30 '20 at 18:04
  • @user140242, If $p=2$ then from the condition $a

    – ASP Oct 01 '20 at 06:56
  • @DavidJones Then you have to edit the post where you say "where $ a$ and $m>1$" – user140242 Oct 01 '20 at 07:01
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    @user140242 I understood it as "$a$ and $m$ are positive integers and $m>1$" and not "$a>1$ and $m>1$ are positive integers". – supinf Oct 01 '20 at 08:38
  • [1]: **Fact**: If $n=\prod p_i^{t_i}$ is not square free, then $\gcd(\phi(n),n-1) < \dfrac{n}{p}$; where $p$ is the least prime divisor of $n$. **Proof**: Note that $1\leq\dfrac{p_j}{p}$. Also we have $\gcd(p_i^{t_i-1}, n-1)=1$, so we can conclude that $\gcd(\phi(n),n-1)$ $=\gcd(\prod (p_i-1),n-1)$ $\leq\prod (p_i-1)$ $<\prod p_i$ $\leq\dfrac{p_j\prod p_i}{p}$ $\leq\dfrac{n}p$. (For the last inequality, note that the numerator, $p_j\prod p_i$, divides $n$.)$\Box$ – Davood Oct 01 '20 at 21:52
  • [2]: Returning back to your question: If $n$ is not prime (equivalently $-\dfrac{n}{p}<-1$), and if we also assume another extra assumtion that "$n$ is not square free", then your statement is true. **Proof**: Suppose on contrary there exist $m$ and $a$ with desired properties. Clearly $m \mid \gcd(\phi(n),n-1)$, by considering the fact in "the comment [1]", we have $m<\dfrac{n}{p}$. Also notice that $a \leq (p-1)$. So we have $n-1=am<(p-1)(\dfrac{n}{p})=n-\dfrac{n}{p} – Davood Oct 01 '20 at 21:53
  • [3]: I think the fact in "the comment [1]" is true for an arbitrary composite natural number $n$, but I don't have any idea how should I prove it for the case of square-free numbers $n=p_1 \dots p_k$. If Either of these two (My fact, or your question) is wrong, then the counter-example should be a square-free number, but I could not find a counter-example. – Davood Oct 01 '20 at 22:01
  • [4] I apologize! I misread the question! You wrote that $p$ is the least prime divisor of $m$, but I assumed that "$p$ is the least prime divisor of $n$" **wrongly**! I find a counter-example for **my question —based upon my wrong assumption—**! Let $n=2701=37\times73$, then $p=37$ is the least prime divisor of $n$. Let $a=25<37=p$, and $m=108$. – Davood Oct 01 '20 at 22:50
  • Note this is related to the unsolved [Lehmer's totient problem](https://en.wikipedia.org/wiki/Lehmer%27s_totient_problem), which asks if "... there is any composite number $n$ such that Euler's totient function $\varphi(n)$ divides $n-1$". If there's no satisfactory answer sometime after the bounty expires, based on my limited knowledge & experience with MathOverflow, I believe this meets their minimum requirements. Nonetheless, please be aware others there may not agree with my assessment, so you should first check their help, & at least a few questions, to decide for yourself before posting. – John Omielan Oct 01 '20 at 23:15
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    One other thing is that if you do later post your question on [MathOverflow](https://mathoverflow.net/), please make sure to add a link in your question text here to that other question, and in the MathOverflow question back to this one. This will help to avoid people duplicating efforts on one site which has already been done on the other site. – John Omielan Oct 02 '20 at 03:05
  • @John Omielan. Am not sure about the relationship of this conjecture with Lehmer's totient problem. From my investigation a proof of this conjecture leads to a generalisation( an extension) of Lucas's converse of Fermat's little theorem. – ASP Oct 02 '20 at 07:22
  • @DavidJones I believe I should've written "somewhat related" instead of "related". For any conjecture counter-example, with $m \mid \phi(n) \implies \phi(n) = km, \; k \ge 1$, if $k \mid a$, then $\phi(n) \mid n-1$ meaning it'll solve the Lehmer's totient problem. Thus, it's much more likely $k \not\mid a$ and, in particular, $k \gt 2$, so this is possibly another restriction on any counter-example. Of course, even if the problems are somewhat related, it doesn't mean that any methods to solve them, use or consequences of the solution, etc., have much, if anything, in common with each other. – John Omielan Oct 02 '20 at 17:55
  • There are no counterexamples with $m \le 10^{10}$ – ASP Jan 24 '21 at 11:22

4 Answers4


Partial answer:

Lemma: Let $n=am+1$ where $a\ge1$ and $m\ge2$ are integers. Suppose that $m\mid\phi(n)$ and $a<p$ where $p=\min\{p^*\in\Bbb P:p^*\mid m\}$. If $n$ is not prime then either

  • $n$ is of the form $\prod p_i$ where $p_i$ are primes, or

  • $n$ is of the form $2^kr$ where $k,r$ are positive integers.

Proof: Suppose that $n$ is composite. First, note that $m$ must be odd as otherwise, $a=1$ which yields $n-1=m$. The condition $m\mid\phi(n)$ forces $n$ to be prime which is a contradiction.

Next, write $n=q^kr$ where $k,r$ are positive integers and $q$ is a prime such that $(q,r)=1$. As $\phi(n)=q^{k-1}(q-1)\phi(r)$ the condition $m\mid\phi(n)$ yields $$q^{k-1}(q-1)\phi(r)=mt\implies aq^{k-1}(q-1)\phi(r)=t(q^kr-1)$$ for some positive integer $t$. It follows that either $k=1$ or $t=q^{k-1}v$ for some integer $v\ne t$. In the latter case, we obtain $$\frac{q^kr-1}{q^{k-1}(q-1)\phi(r)}=\frac{aps}{mt}=\frac at\implies p>\frac{t(q^kr-1)}{q^{k-1}(q-1)\phi(r)}.$$ Combining this with the trivial result $p<q^{k-1}(q-1)\phi(r)/t$ yields $$t<\frac{q^{k-1}(q-1)\phi(r)}{\sqrt{q^kr-1}}\implies v<\frac{(q-1)\phi(r)}{\sqrt{q^kr-1}}.$$ Substituting back into $n=am+1$ gives $$q^kr-1=\frac av(q-1)\phi(r)\implies aq\phi(r)-vq^kr=a\phi(r)-v>\phi(r)\left(a-\frac{q-1}{\sqrt{q^kr-1}}\right)$$ which is positive since $k\ge2$. This yields $a>vq^{k-1}\ge vq$. Since $p$ is the least prime divisor of $m$, we have $p\le q-1$, unless $q=2$ or $q-1=v$.

Evidently, the first case contradicts $a<p$, so $k=1$. This means that $n$ must be of the form $\prod p_i$ where $p_i$ are primes. The condition $m\mid\phi(n)$ gives $\prod(p_i-1)=bm$ for some positive integer $b$, and substituting this into $n=am+1$ yields $$a=b\frac{\prod p_i-1}{\prod(p_i-1)}.$$ When $m$ is even, we have $a<p\implies a<2$ which implies that $m=\prod p_i-1$. Further, $$b<\frac{2\prod(p_i-1)}{\prod p_i-1}<2\implies m=\prod(p_i-1).$$ The only way that $\prod p_i-1=\prod(p_i-1)$ is when $\prod p_i$ is prime, which solves the problem. Finally, notice that $m$ is odd only when $b=2^{\nu_2(\prod(p_i-1))}d$ for some positive integer $d$, so the condition $a<p$ yields $$2^{\nu_2(\prod(p_i-1))}d\frac{\prod p_i-1}{\prod(p_i-1)}<\frac{p_j-1}{2^{\nu_2(p_j-1)}}$$ for some prime $p_j\mid\prod p_i$.

The second case $q=2$ implies that $n=2^kr=am+1$ where $m\mid\phi(r)$; that is, for some positive integer $g$ we have $g(2^kr-1)=a\phi(r)$.

The third case $q-1=v$ forces $m=\phi(r)$, so $m=1$. This is a contradiction as there is no prime $p$ that can divide $m$.

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    I've read through the rest of your answer and have $2$ more questions. First, with "... since $p$ is the least prime divisor of $m$, we also have $p\le q-1$ ...", there's the possibility that $p \mid \phi(r)$. What you've shown is that there's no prime factor $q \gt p$ which can have more than one factor in $n$, but I don't see how this excludes possibly smaller factors than $p$ occurring more than once. The second issue is in your final inequality, I don't understand what your $p_c$ is. – John Omielan Oct 03 '20 at 20:16
  • I don't understand what your statement "The smallest divisor of $ab/c$ is at most $a$ (under trivial constraints)" has to do with my comment, e.g., since the smaller $p_i \mid n = am + 1$ have no direct connection with $a$ or $m$. Thanks for clarifying your $p_c$, now called $p_j$. This seems to be one of the primes where $p \mid p_j - 1$, so you may wish to make this clear. Also, thanks for providing feedback in your multiple replies. However, I've spent more time working with your answer than I want to, so please understand why if I don't respond to any additional replies from you. – John Omielan Oct 04 '20 at 03:43
  • @JohnOmielan Remember that $p$ divides $m$ not $n$. But your comment is useful as you have pointed out the flaw: the case $q=2$ means that $m=\phi(r)/v$ so $p$ cannot be less than $q-1$. Notice that this case is similar to the $n=\prod p_i$ case as the integer $g$ as updated in my answer must absorb all the powers of $2$ just like with $b$. – TheSimpliFire Oct 04 '20 at 10:01
  • @TheSimpliFire, from the statement of your Lemma, are the $p_i$ all odd primes? Secondly are they distinct?. If they are not distinct then all your lemma says is if $n$ is not prime then $n$ is either odd or even something which is true for all positive integers. How does this contribute to the required proof? – ASP Oct 04 '20 at 11:19
  • The $p_i$ are all odd as $a=2^{\nu_2(\prod(p_i-1))}d(\prod p_i-1)/\prod(p_i-1)$ is even, so necessarily $n=am+1$ is odd. The $p_i$ are all distinct as $k=1$. We need to eliminate the two cases $n=\prod p_i$ and $n=2^kr$ to complete the proof. – TheSimpliFire Oct 04 '20 at 11:36
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    In your proof to show that $k=1$, you showed that $p \le q-1$ to arrive at a contradiction. However there's a fatal error in the proof. Since $p$ is a divisor of $m$ and $m \ | \ \phi(n) $ we have $p \ | \ \phi(n)=q^{k-1}(q-1) \phi(r) $. Because $p$ is prime, $p$ divides at least one of the factors : $q$, $q-1$ or $ \phi(r) $. In your proof you have assumed that $p$ divides the factor $q-1$ to arrive at $p \le q-1$. You haven't considered the case when $p \ | \ \phi(r) $ – ASP Oct 04 '20 at 13:10
  • It doesn't matter; as long as $q-1\ge2$ the *smallest* prime divisor is always less than or equal to either $q-1$ or $\phi(r)$. For example, if the smallest prime divisor $p$ divides $ab$ where $a,b>1$ then $p\le a$ is valid, and so is $p\le b$. – TheSimpliFire Oct 04 '20 at 13:14
  • @TheSimpliFire all you have said is right. Now take the case when $p \ | \ \phi(r) $, $p \le \phi(r) $. Can you still arrive at the contradiction $a>p$? – ASP Oct 04 '20 at 13:40
  • We only need to consider $p\mid\phi(r)$ when $q=2$ (since $q-1=1$ has no prime divisors) or when $q-1=v$. I have considered both cases in the last two paragraphs of my answer. The case $q=2$ is still unsolved. Otherwise, why make life difficult when we can take $p\le q-1$ rather than $p\le\phi(r)$ when both are just as valid? – TheSimpliFire Oct 04 '20 at 13:41
  • If $p \ | \ ab$ then either $p \ | \ a$ or $ p \ | \ b$ not $p \ | \ a$ and $p \ | \ b$. Take $p=3$, $ q-1=4>2$ and $\phi(r) =12$. Do you see that $p=3 \ | \ \phi(r) =12$ but $p=3 \nmid q-1=4$? – ASP Oct 04 '20 at 14:06
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    That is not a correct example. Remember that $p$ is the ***least*** prime divisor. So when we have $q-1=4$ and $\phi(r)=12$, the correct $p$ is $2$, and evidently $p\le q-1$ ***and*** $p\le\phi(r)$. I am ***not*** saying that $p\mid a$ and $p\mid b$. I am saying that $p\le a$ and $p\le b$. I hope this is clearer now. – TheSimpliFire Oct 04 '20 at 14:11


First, let the prime factorization of $m$ and $n=am+1$ be: $$m=\prod_{i=1}^k p_i^{a_i} \quad \quad \quad n=\prod_{i=1}^l q_i^{b_i}$$ where $p_1$ is the least prime factor of $m$. Since $\gcd(m,am+1)=1$, all $p_i$'s and $q_i$'s are pairwise distinct. Using this, we have: $$m \mid \phi(n) \implies \prod_{i=1}^k p_i^{a_i} \mid \prod_{i=1}^l(q_j-1)q_j^{b_j-1} \implies \prod_{i=1}^k p_i^{a_i} \mid \prod_{i=1}^l(q_i-1)$$ If there exists a prime $q_j>p_1$ such that $\gcd(m,q_j-1)$, then we would have: $$\phi(am+1) \geqslant \prod_{i=1}^k (q_i-1) \geqslant (q_j-1)m \geqslant p_1m$$ which is a contradiction. We also arrive at a similar contradiction if we assume that $b_j>1$ for any $q_j>p_1$. Thus, we can conclude that: $$am+1=M\prod_{i=1}^s r_i$$ where $r_i>p_1$ are primes and $M$ has all prime factors less than $p_1$. As we know that $m \mid \prod (r_i-1)$, it follows that we have $am+1 > Mm$. Thus, $p_1 > a \geqslant M$. If there exists a prime $p_j \mid m$, such that $p_j^{a_j+1} \mid \phi(n)$, then: $$\phi(am+1) \geqslant p_jm \geqslant p_1m > am+1$$ which is obviously a contradiction. Thus, we must have $p_j^{a_j} \mid \mid \phi(n)$ and as a consequence, $s \leqslant \sum a_i$. We can solve particular cases using these facts.

The case $m=p^t$

When $m$ is a perfect prime power, we can take $m$ to be odd. We must have $r_i \equiv 1 \pmod{p}$. We know that we have $p^t \mid \mid \prod (r_i-1)$. The equation becomes: $$ap^t+1 = M\prod_{i=1}^s r_i \implies M \equiv 1 \pmod{p}$$ Since $M<p$ this forces $M=1$. Next, we can write $r_i=p^{b_i}Q_i+1$ where $p \nmid Q_i$. We know that $\sum b_i = t$. $$ap^t+1 = \prod_{i=1}^s (p^{b_i}Q_i+1) \implies ap^t > p^t \cdot \prod Q_i \implies a > \prod_{i=1}^s Q_i$$ The strict inequality is ensured since $s>1$ i.e. $n$ is not prime. WLOG assume $b_1 \leqslant b_2 \leqslant \cdots \leqslant b_s$. Let $c=b_1=b_2=\cdots = b_x<b_{x+1}$. Taking the equation modulo $p^{c+1}$ gives: $$p^c\sum_{i=1}^x Q_i \equiv 0 \pmod{p^{c+1}} \implies p \mid \sum_{i=1}^x Q_i \implies \sum_{i=1}^x Q_i>a>\prod_{i=1}^x Q_i$$ However, since all $r_i$ are odd, all $Q_i$ must be even (since $p$ is odd). This would yield a contradiction since all $Q_i > 1$ and thus, the above inequality of sum being greater than product cannot hold. Thus, $n$ cannot be composite.

The case $m=pq$

Subcase $1$ : $s=1$ $$apq+1=Mr$$ Since $pq \mid (r-1)$, we have $M \equiv 1 \pmod{pq}$ and thus, $M=1$. However, this gives $n=Mr=r$ which is prime.

Subcase $2$ : $s=2$ $$apq+1=Mr_1r_2$$ Let $p \mid (r_1-1)$ and $q \mid (r_2-1)$. Moreover, let $p<q$. Writing $r_1=pQ_1+1$ and $r_2=qQ_2+1$ gives: $$apq+1=M(pqQ_1Q_2+pQ_1+qQ_2+1) \implies (a-MQ_1Q_2)pq+1=M(pQ_1+qQ_2+1)$$ Since the RHS is positive, this gives $a-MQ_1Q_2 \geqslant 1$. We have: $$pq < MQ_1Q_2 \bigg(\frac{p}{Q_2}+\frac{q}{Q_1}+\frac{1}{Q_1Q_2}\bigg) \implies q < \frac{p+1}{Q_2}+\frac{q}{Q_1} < \frac{q}{Q_1}+\frac{q}{Q_2} \leqslant q$$ This is a contradiction. Thus, $n$ cannot be composite.

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Let $n=am+1, m|φ(n), a,m>1, a<p, p$ is the least factor of $m$.

Let $n$ be a composite number with prime factorization

$$n=p_1^{e_1} p_2^{e_2 }\dots p_k^{e_k}$$

Without loss of generality, let $p_1 \lt p_2 \lt \dots < p_k$.

$$φ(n)=n(1-{1 \over p_1} )(1-{1 \over p_2} )…(1-{ 1 \over p_k} )$$

$$=p_1^{e_1} p_2^{e_2}\dots p_k^{e_k} {(p_1-1) \over p_1 } {(p_2-1) \over p_2 }…{(p_k-1) \over p_k }$$

$$=p_1^{e_1-1} p_2^{e_2-1} \dots p_k^{e_k-1} (p_1-1)(p_2-1)…(p_k-1)$$

Since $m | φ(n)$, we can write for some integer $t$,

$$φ(n)=mt=p_1^{e_1-1} p_2^{e_2-1}\dots p_k^{e_k-1} (p_1-1)(p_2-1) \dots (p_k-1)$$

$$⇒m= {(p_1^{e_1-1} p_2^{e_2-1}…p_k^{e_k-1} (p_1-1)(p_2-1)…(p_k-1)) \over t}$$

The terms $(p_2-1),…,(p_k-1)$ in the numerator are all even since $p_2,…,p_k$ are primes. For the case of $p_1 = 2$, $p_1-1 = 1$.

We can write for integer $r_1, r_2, \dots, r_k$,

$$m={ p_1^{e_1-1} p_2^{e_2-1} \dots p_k^{e_k-1} r_1 r_2…r_k 2^k \over t}$$

$t$ must be of the form $2^k c$ where $c$ divides $p_1^{e_1-1} p_2^{e_2-1}\dots p_k^{e_k-1} r_1 r_2 \dots r_k$. Also note that if $p_1$ is 2, $p_1^{e_1-1}$ must be a factor of $c$. Otherwise the least factor of $m$ will be 2 and $p = 2$ which causes $a = 1$ since $a<p$ by definition. However, $a>1$ by definition.

$$m={p_1^{e_1-1} p_2^{e_2-1} \dots p_k^{e_k-1} r_1 r_2 \dots r_k \over c}$$

$$n=am+1=a{p_1^{e_1-1} p_2^{e_2-1}…p_k^{e_k-1} r_1 r_2…r_k \over c}+1$$

By definition, $p$ is the least divisor of $m$. The maximum value that $p$ can take is $p_k$ since $r_j<p_k,∀ 1≤j≤k$. By definition, $a<p$. Note that $c$ will have common factors with $a{ p_1^{e_1-1} p_2^{e_2-1} \dots p_k^{e_k-1} r_1 r_2…r_k 2^k}$, but cannot be exactly ${ p_1^{e_1-1} p_2^{e_2-1} \dots p_k^{e_k-1} r_1 r_2…r_k 2^k}$. If it were the case, $m = 1$ which conflicts with the assumption $m>1$. So, the factors of $c$ must have at most $e_j - 1$ exponent for the prime factor $p_j$ for all $1 \le j \le k$.

So, we have

$$n=p_1^{e_1 } p_2^{e_2 } \dots p_k^{e_k} = a{p_1^{e_1-1} p_2^{e_2-1} \dots p_k^{e_k-1} r_1 r_2…r_k \over c}+1$$

Let $p_u$ be the smallest prime that is the common factor of ${p_1^{e_1-1} p_2^{e_2-1} \dots p_k^{e_k-1} r_1 r_2…r_k \over c}$ and $n$. $p_u$ exists since we have proved that the maximum exponent of prime factor $p_j$ of $c$ is less than $e_j - 1$.

Taking modulo $p_u$, we get

$$0≡1 \mod p_u$$

This is impossible. Therefore $n$ must be prime.

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    Why does such a $p_u$ exist? Maybe it could happen that they have no common factor? – supinf Oct 06 '20 at 18:15
  • Also it is not clear to me why $p_1-1$ should be even. – supinf Oct 06 '20 at 18:17
  • @supinf: Please see edits made above. $p_u$ exists because $c$ cannot be exactly the numerator as it would cause $m = 1$. $p_1 -1$ does not have to be even, but that happens only when $p_1$ is 2 and $p_1 - 1$ is $1$ and $r_1 = 1$. If $p_1$ is 2, $p_1^{e_1-1}\times 2^k$ must be a factor of $c$ and the rest of the proof stays intact. – vvg Oct 07 '20 at 02:57
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    Your assumption that $p \le p_k$ is wrong. take $n=175=6 \bullet 29 +1$ with $a=6$, $m=29$ and $p=29$.The largest prime divisor of $n=p_k=7$. Clearly $p>p_k$. – ASP Oct 07 '20 at 04:37
  • @davidjones: The definition of $p$ is it is the least prime divisor of $m$ (not $n$). $p_k$ is the largest prime divisor of $m$. So,$p \le p_k$. – vvg Oct 07 '20 at 04:45
  • .. $p_k$ is largest prime divisor of $n$ – vvg Oct 07 '20 at 04:52

As mentioned if $2 \mid m$ then $p=2$ and $a=1$.

So we consider the case $m$ odd and $p \ge 3$

$m= \prod_\limits{i=1}^k q_i^{a_i}$ with $q_i$ prime and $p=q_1<q_2< \dots <q_k$

$n=1+am=\prod_\limits{j=1}^r p_j^{k_j}$ with $p_j$ prime

we take $p_1$ so that $q_1 \mid (p_1-1)$

$\phi (n)=\prod_\limits{j=1}^r p_j^{k_j-1}(p_j-1)=bm$

$m=\frac{\prod_\limits{j=1}^r p_j^{k_j-1}(p_j-1)}{b}$

if $p_j \mid m$ then $n=1 \bmod p_j$ therefore it must be $p_j \not \mid m$

then $b=c\prod_\limits{j=1}^r p_j^{k_j-1}$ , $c \ge 1$

$n=1+am=1+a \frac{\prod_\limits{j=1}^r p_j^{k_j-1}(p_j-1)}{b}=\prod_\limits{j=1}^r p_j^{k_j}$


$ a=b \frac{\prod p_j^{k_j}-\displaystyle 1}{\prod p_j^{k_j-1}(p_j-1)}=c\prod p_j^{k_j-1} \frac{\prod p_j^{k_j}-\displaystyle 1}{\prod p_j^{k_j-1}(p_j-1)}$

but $ \frac{\prod p_j^{k_j}-\displaystyle 1}{\prod p_j^{k_j-1}(p_j-1)}>1$

then if exists $k_j \geq 2 $ for some $j$ with $q_i \mid (p_j-1)$

$a>p_j>(p_j-1)>q_i \geq q_1=p$

in this case $n$ composite number and $a>p$

therefore we have to analyze the cases in which $k_j =1$ for each $j$ with $q_i \mid (p_j-1)$


$$n=n_1 \prod_\limits{j=1}^t p_j \text{ where some } q_i \mid (p_j-1) \text{ and } n_1 \ge 1 \text{ with } q_i \not\mid \phi (n_1)$$

Case 1

$m \mid \phi(p_1)$ then $m \mid (p_1-1)$

$p_1=1+fm$ , $f \ge 2$

as demonstrated in the case $n=1+apq$

$n=1+am=ep_1$ with $e \ge 1$

$e(fm + 1) = am + 1 $

$(ef)m + e = am + 1 $

$e - 1 = (a - ef)m$

then $m \mid e - 1$, but $m \gt a \ge ef$ so $e \lt m$, then $e = 1$

therefore in this case it must be $n = p_1$ prime and $am = \phi(p_1)=fm$ .

Case 2

$m \not\mid (p_1-1)$



$q_1d_1 \mid (p_1-1)$ , $d_1 \ge 1$

$d_2 \mid \phi(p_2)$ , $d_2 > q_1$

$d_2 \not\mid (p_1-1)$

$p_1=1+fq_1d_1$ , $f \ge 2$




then $q_1d_1 \mid (p_2-1)$

$p_2=1+gq_1d_1$ , $g \ge 2$


$a= \frac{f+g+fgq_1d_1}{d_2}$

but $\phi (p_2)=gd_1q_1$ and $d_2 \mid gd_1q_1$

but $d_2 \not\mid (p_1-1)$ then $d_2 \not\mid q_1d_1$ and $d_2 \mid g$

then $d_2 < g$ and $a= \frac{f+g+fgq_1d_1}{d_2}>\frac{fgq_1d_1}{d_2}>\frac{fgq_1d_1}{g}>fq_1d_1>q_1$

in this case $n$ composite number and $a>p$

Case 3

$m \not\mid (p_1-1)$

$n=1+am=n_1 \prod_\limits{j=1}^t p_j \text{ where for each p_j there is some } q_i \text{ with } q_i\mid (p_j-1)$

$q_i \not\mid \phi (n_1)$

$p_j=1+b_jq_{i_1}\dots q_{i_{m_j}}$ with $b_j\ge 2$


$\phi(n)=\phi(n_1) \prod(p_j-1)=\phi(n_1) \prod (b_jq_{i_1}\dots q_{i_{m_j}})=cbm$

but $\frac{n-1}{\phi(n)}=\frac{am}{cbm}>1$

then $cb= \phi(n_1) \prod b_j <a$ with $b \ge 2^{a_1+\dots +a_k}$

but $n=n_1 \prod_\limits{j=1}^t(1+b_jq_{i_1}\dots q_{i_{m_j}})=1+am$

and $\prod_\limits{j=1}^t(1+b_jq_{i_1}\dots q_{i_{m_j}})=(1+b_1q_{i_1}\dots q_{i_{m_1}}+\dots+\prod_\limits{j=2}^t(b_jq_{i_1}\dots q_{i_{m_j}})+\prod_\limits{j=1}^t(b_jq_{i_1}\dots q_{i_{m_j}}))$

then $m(a-n_1b)=-1+n_1(1+b_1q_{i_1}\dots q_{i_{m_1}}+\dots+\prod_\limits{j=2}^t(b_jq_{i_1}\dots q_{i_{m_j}}))$

from which $(a-n_1b)> 0$

then $a>n_1b$

$m<-1+n_1(1+b_1q_{i_1}\dots q_{i_{m_1}}+\dots+\prod_\limits{j=2}^t(b_jq_{i_1}\dots q_{i_{m_j}})$

but each addend $\prod(b_jq_{i_1}\dots q_{i_{m_j}})<\frac{bm}{q_{i_k} \prod b_k}<\frac{bm}{q_12^{s_k}}$ with $q_{i_k} \not=q_{i_1}\not=\dots\not= q_{i_{m_j}}$ and $ b_k \not= b_j$

then $m<\frac{n_1bm}{2q_1}(1+\sum\frac{1}{2^{s_k-1}})<\frac{n_1bm}{q_1}<\frac{am}{q_1}$

therefore $a>q_1$ and this contradicts the initial condition $a<q_1$

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