When is $\frac{5^n - 1}{4}$ squarefree?

Question

While researching the topic of odd perfect numbers, I came across the following related subproblem:

PROBLEM: Determine congruence conditions (on $n > 1$) such that $$\frac{5^n - 1}{4}$$ is squarefree.

MY ATTEMPT

Set $$m_n := \frac{5^n - 1}{4}.$$

I noticed that $m = m_1 = 1$ is squarefree for $n = 1$.

Next, I considered the prime factorizations of $m_n$ for the first fifteen values (i.e., $2 \leq n \leq 16$): $$\begin{array}{|c|c|c|} \hline \text{Value of } n &\text{Value of } m_n & \text{Prime Factorization of } m_n \\ \hline 2 & 6 & 2 \times 3 \\ \hline 3 & 31 & 31 \\ \hline 4 & 156 & {2}^2 \times 3 \times 13 \\ \hline 5 & 781 & 11 \times 71 \\ \hline 6 & 3906 & 2 \times {3}^2 \times 7 \times 31 \\ \hline 7 & 19531 & 19531 \\ \hline 8 & 97656 & {2^3} \times 3 \times 13 \times 313 \\ \hline 9 & 488281 & 19 \times 31 \times 829 \\ \hline 10 & 2441406 & 2 \times 3 \times 11 \times 71 \times 521 \\ \hline 11 & 12207031 & 12207031 \\ \hline 12 & 61035156 & 2^2 \times 3^2 \times 7 \times 13 \times 31 \times 601 \\ \hline 13 & 305175781 & 305175781 \\ \hline 14 & 1525878906 & 2 \times 3 \times 29 \times 449 \times 19531 \\ \hline 15 & 7629394531 & 11 \times 31 \times 71 \times 181 \times 1741 \\ \hline 16 & 38146972656 & 2^4 \times 3 \times 13 \times 17 \times 313 \times 11489 \\ \hline \end{array}$$

From this initial data sample, I predict the truth of the following conjectures:

CONJECTURE 1: If $n \equiv 0 \pmod 6$, then $m_n$ is not squarefree.
CONJECTURE 2: If $n \equiv 0 \pmod 5$, then $m_n$ is squarefree.
CONJECTURE 3: If $n \equiv 0 \pmod 3$ and $n$ is odd, then $m_n$ is squarefree.
OBSERVATION - If $m_n$ is a prime number, then $n$ must also prime.

I skimmed through OEIS sequence A005117 and did not find any references to these conjectures.

ATTEMPTING TO RESOLVE CONJECTURE 1

I searched for counterexamples to Conjecture 1 using Pari-GP in Sage Cell Server, I did not get any output in the range $2 \leq n \leq 100$:

This gives further computational evidence for Conjecture 1.

RESOLVING CONJECTURE 2

I searched for counterexamples to Conjecture 2 using Pari-GP in Sage Cell Server, I got the following output in the range $2 \leq n \leq 100$:

20[2, 2; 3, 1; 11, 1; 13, 1; 41, 1; 71, 1; 521, 1; 9161, 1]
30[2, 1; 3, 2; 7, 1; 11, 1; 31, 1; 61, 1; 71, 1; 181, 1; 521, 1; 1741, 1; 7621, 1]
40[2, 3; 3, 1; 11, 1; 13, 1; 41, 1; 71, 1; 241, 1; 313, 1; 521, 1; 9161, 1; 632133361, 1]
55[11, 2; 71, 1; 103511, 1; 511831, 1; 12207031, 1; 65628751, 1; 190295821, 1]
60[2, 2; 3, 2; 7, 1; 11, 1; 13, 1; 31, 1; 41, 1; 61, 1; 71, 1; 181, 1; 521, 1; 601, 1; 1741, 1; 2281, 1; 7621, 1; 9161, 1; 69566521, 1]
80[2, 4; 3, 1; 11, 1; 13, 1; 17, 1; 41, 1; 71, 1; 241, 1; 313, 1; 521, 1; 9161, 1; 11489, 1; 25601, 1; 632133361, 1; 909456847814334401, 1]
90[2, 1; 3, 3; 7, 1; 11, 1; 19, 1; 31, 1; 61, 1; 71, 1; 181, 1; 521, 1; 829, 1; 1171, 1; 1741, 1; 5167, 1; 7621, 1; 169831, 1; 297315901, 1; 60081451169922001, 1]
100[2, 2; 3, 1; 11, 1; 13, 1; 41, 1; 71, 1; 101, 1; 251, 1; 401, 1; 521, 1; 1901, 1; 9161, 1; 239201, 1; 9384251, 1; 424256201, 1; 50150933101, 1; 89620825374601, 1]

This means that Conjecture 2 is false.

RESOLVING CONJECTURE 3

I searched for counterexamples to Conjecture 3 using Pari-GP in Sage Cell Server, I got the following output in the range $2 \leq n \leq 100$:

93[31, 2; 1861, 1; 148429, 1; 878851, 1; 625552508473588471, 1; 172974812463239310024750410929, 1]

This means that Conjecture 3 is false.

RESOLVING AN OBSERVATION

I searched for counterexamples to Conjecture 4 using Pari-GP in Sage Cell Server, I did not get any output in the range $2 \leq n \leq 888$.

I did search for the primes $n$ satisfying Conjecture 4, in the range $2 \leq n \leq 888$, here is the output:

3Mat([31, 1]) 
7Mat([19531, 1]) 
11Mat([12207031, 1]) 
13Mat([305175781, 1]) 
47Mat([177635683940025046467781066894531, 1])

127Mat([14693679385278593849609206715278070972733319459651094018859396328480215743184089660644531, 1])
149Mat([35032461608120426773093239582247903282006548546912894293926707097244777067146515037165954709053039550781, 1])
181Mat([815663058499815565838786763657068444462645532258620818469829556933715405574685778402862015856733535201783524826169013977050781, 1])
619Mat([1149139339972917612905859963756803959098341443304966913142347324923898626737719076860276030395228900386874376453787094334535897724625161448243286016083472616301829493727589186752415411518815513513188653607789335439056683481774870999044049378038610542097905774057044077931436914036888020005064547897035207825993485927279302571673236827045083340815370132092682949566420825990422295705880874452974385602210816159640671685338020324707031, 1])

This means that the numbers $$\frac{5^3 - 1}{4}, \frac{5^7 - 1}{4}, \frac{5^{11} - 1}{4}, \frac{5^{13} - 1}{4},$$ $$\frac{5^{47} - 1}{4}, \frac{5^{127} - 1}{4}, \frac{5^{149} - 1}{4}, \frac{5^{181} - 1}{4}, \frac{5^{619} - 1}{4}$$ are all prime.

We end this section with the following prediction:

CONJECTURE 4: The divisibility condition $n \mid m_n$ holds when either (a) $5 \nmid n$; or (b) $n$ is an odd prime.

Alas, this where I get stuck, as I do not currently know how to prove Conjectures 1 and 4.

I have removed some tags since the question as stated has nothing to do with them. Many elementary questions arise in numerous advanced contexts but that doesn't mean that their solution has anything to do with these advanced contexts. — Bill Dubuque, May 14 '23 at 20:13
Note that conjectures 1 and 2 are mutually exclusive, as witnessed by all $n$ congruent to $0$ mod $30$. — Servaes, May 14 '23 at 22:12
Thanks, @Servaes! Would you care to add an answer, with more details please? Also, I wonder what can be said about $m_n$ when $n$ is restricted to the appropriate residue classes modulo $60$? — Jose Arnaldo Bebita Dris, May 15 '23 at 01:07
@JoseArnaldoBebitaDris There are no details really; conjecture 2 claims that $m_{30}$ is squarefree, conjecture 1 claims that $m_{30}$ is not squarefree. — Servaes, May 15 '23 at 06:55
I agree. So does that argument for $30$ "lift" to $60$, @Servaes? =) — Jose Arnaldo Bebita Dris, May 15 '23 at 18:45
@JoseArnaldoBebitaDris With all due respect I fail to see what the issue is here, and wonder if you are just being sarcastic or something? This should be blindingly obvious. The same is true for any multiple of both $5$ and $6$, i.e. any multiple of $30$. — Servaes, May 15 '23 at 19:32
@Servaes: Ohh, I was not being sarcastic. I was referring to the sexagesimal system. — Jose Arnaldo Bebita Dris, May 15 '23 at 20:02
@Servaes: That being said, since you do confirm that the argument for $30$ lifts to $60$, can we just both "agree to disagree"? =) — Jose Arnaldo Bebita Dris, May 15 '23 at 20:04
@JoseArnaldoBebitaDris I still think you're pulling my leg, so I wont respond further. — Servaes, May 15 '23 at 20:19
@Servaes: Seriously? Consider this and then this. Now pray, please tell me, am I still "pulling your leg"? =) — Jose Arnaldo Bebita Dris, May 15 '23 at 20:35
@JoseArnaldoBebitaDris I have no idea what to make of these two webpages. I am very tired of your purposely unclear responses. — Servaes, May 15 '23 at 20:37
OEIS sequence A004061 lists $$3, 7, 11, 13, 47, 127, 149, 181, 619, 929, 3407, 10949, 13241, 13873, 16519, 201359, 396413, 1888279, 3300593$$ as the sequence of exponents $k$ such that $(5^k - 1)/4$ is prime. — Jose Arnaldo Bebita Dris, May 28 '23 at 03:55
@BillDubuque: Further to your first comment to this question here dated May 14, 2023, I highly suggest that you do not touch the current tags for this question, in view of this preprint. (By the way, I am sure that Bill Dubuque is not even your real name.) =) — Jose Arnaldo Bebita Dris, May 28 '23 at 04:03
@Jose As I explained before, simple results in elementary number theory often have applications in many different fields. But that does not imply that one should tag it in those fields. Note that none of the answers say a word about those tags you added. Indeed the proofs have nothing to do with them. — Bill Dubuque, May 28 '23 at 04:21
@BillDubuque: Seriously? Can you honestly say that? For instance, can you please explain (in layman's terms) why these proofs (as you say) "have nothing to do with them"? — Jose Arnaldo Bebita Dris, May 28 '23 at 04:46
@BD: Further to your last comment, note that my "unspoken" premise is that this question is about/related to odd perfect numbers. But did I not mention that, at all, in the beginning of my post? =) — Jose Arnaldo Bebita Dris, May 28 '23 at 04:50
@BD: That being said, I believe (and I will only say this once) that you are specifically targetting my posts. I will take the matter up in Meta MSE. See you there! =) — Jose Arnaldo Bebita Dris, May 28 '23 at 04:53
@Jose Yes, seriously. There is no mention of divisor-sums, perfect-numbers or pseudoprimes in any of the answers. I correct the tags of many posts. Yours are but a very small fraction. — Bill Dubuque, May 28 '23 at 04:53
@BD: You must have corrected the tags of a lot of questions. =) — Jose Arnaldo Bebita Dris, May 28 '23 at 04:56
@BD: Now pray, tell me, which among my questions did you remove/replace my originally-specified tags on? — Jose Arnaldo Bebita Dris, May 28 '23 at 04:58

score 4 · Accepted Answer · answered May 14 '23 at 19:10

Conjectures 2 and 3 are false, as shown by the counterexamples $$ \frac{5^{55}-1}4 = 11^2\times 71\times 103511\times 511831\times 12207031\times 65628751\times 190295821 $$ and $$ \frac{5^{93}-1}4 = 31^2\times 1861\times 148429\times 878851\times 625552508473588471\times 172974812463239310024750410929. $$

The counterexample to Conjecture 2 (for example) was found simply by looking at the primes dividing $\frac{5^{5}-1}4$ and knowing that their squares would eventually appear in the factorizations. For example, $11$ divides $\frac{5^{5}-1}4$, and we can compute that the multiplicative order of $5$ modulo $11^2$ is $55$, which already proves that $\frac{5^{55}-1}4$ is divisible by $11^2$.

Similarly, the multiplicative order of $5$ modulo $71^2$ is $355$, and so $\frac{5^{355}-1}4$ is not squarefree either (it's divisible by $71^2$). This method generalizes to higher powers as well: for example, multiplicative order of $5$ modulo $11^2$ is $605$, which proves that $\frac{5^{605}-1}4$ is not cubefree (it's divisible by $11^3$).

General principle: it's extremely difficult to write down a parametrized family of squarefree numbers; so we should be extremely skeptical when conjecturing that a family of numbers contains only squarefree numbers.

Luca Armstrong · Answer 2 · 2023-05-14T19:31:19.857

3

Here's a proof of conjecture 1:

Suppose that $n \equiv 0 \pmod 6$, meaning $n = 6k$ for some natural number $k$,

$$m_n = \frac{5^{6k} - 1}{4} = \frac{15625^k - 1}{4}.$$

Then observing the numerator, we see that

$$15625 \equiv 1 \pmod {36} \ \ => \ \ 15625^k - 1 \equiv 0 \pmod{36}.$$

Thus, the numerator can be written as $36r$ for some natural number $r$. This means

$$m_n = \frac{36r}{4} = 9r.$$

Finally, $m_n$ is divisible by $9$, hence, not squarefree.

Disproof of conjecture $4$ part $(b)$:

Let $n$ be an odd prime number.

$$n \mid m_n \ <=> \ 4n \mid (5^n - 1)$$

Since $n$ and $4$ are coprime, we have

$$4n \mid (5^n - 1) \ \ <=> \ \ (5^n - 1) \equiv 0 \pmod{4} \ \ \text{and} \ \ (5^n - 1) \equiv 0 \pmod{n}.$$

The first congruence holds trivially, and the second can be reduced using Fermat's Little Theorem to

$$(5^n - 1) \equiv (5 - 1) \equiv 4 \pmod{n}.$$

Thus,

$$4 \equiv 0 \pmod{n} \ => \ n\mid 4$$

However, since $n$ is an odd prime, this can never happen, so we have reached a contradiction.

edited May 14 '23 at 19:31

answered May 14 '23 at 18:45

Luca Armstrong

1,354

Splendid proof, @LucaArmstrong! Would you care taking a stab at Conjecture 4? – Jose Arnaldo Bebita Dris May 14 '23 at 18:46
1

@Jose Arnaldo Bebita Dris you're welcome :), I'll conjecture 4 give a shot. – Luca Armstrong May 14 '23 at 18:48
Thanks again, @LucaArmstrong. Kindly add your comment here to your answer, so that I can upvote and then accept. =) – Jose Arnaldo Bebita Dris May 14 '23 at 19:02
@Jose Arnaldo Bebita Dris Sorry, I made a mistake with my working. The divisibility condition still doesn't hold for all odd primes, but there's no saying it doesn't hold for any odd number $n$. I'll just add this to my answer now – Luca Armstrong May 14 '23 at 19:26
@Jose Arnaldo Bebita Dris I've now updated my answer :) – Luca Armstrong May 14 '23 at 19:32

razivo · Answer 3 · 2023-05-14T19:22:55.647

1

I'll prove a small generalization of conjecture 4. for $p\ne 2,5$, if $p(p-1)|n$ $$\frac{5^n-1}{4}\equiv 0\mod p^2$$ By Euler's theorem $a^{\varphi(n)}\equiv 1\mod n$ for $(a,n)=1$.
We can also notice that $5^n-1|5^{an}-1$.
this implies that if we know the condition for one of $n$'s factors we know it for the whole.

edited May 14 '23 at 19:22

answered May 14 '23 at 19:08

razivo

2,285

Hmm, yes it is not fermat's last theorem. – razivo May 14 '23 at 19:23

When is $\frac{5^n - 1}{4}$ squarefree?

3 Answers3