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

conjectureselementary-number-theorysquarefree-numbers

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.

Best Answer

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.

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