$3^p-2^p$ squarefree?

Question

Original question:

$3^n-2^{n-1}$ seems to be squarefree. Is it?

Answer: No, but amongst the primes dividing one of these numbers, $23$ seems to be a special case: no $23^2$ divide any of them

Are there other case where $p$ divide a $3^n-2^{n-1}$ but $p^2$ does not?

Edited question, based on answers:

It is conjectured that $2^p-1$ is squarefree.

Could it be that $3^p-1$ is also squarefree for $p\neq2$ and $5$ (where $11^2$ appears)?

Could it be that $3^p-2^p$ is also squarefree for $p\neq11$ (where $23^2$ appears)?

Thanks

edit: just saw this Is $(3^p-1)/2$ always squarefree?.

Note: $n$ is a positive natural natural in the original question and $p$ is a prime in my edit.

Answer 1

Well, I posted too quickly.....for $n=67$ it seems we have 2 factor $11$

Answer 2

At the very end, I put a long list of these. The great majority of squared primes are below $100,$ but I did catch a few larger, so far $127, 191, 271, 1021 : \;$

712   +++  = 271^2  cdot mbox{BIG} 
1588  +++ = 73 191^2  cdot mbox{BIG}
2340  +++   = 127^2  cdot mbox{BIG}   
2531  +++     = 1021^2  cdot mbox{BIG} 

There is a way to deal with this without doing genuine factoring. For example, your number is divisible by $49 = 7^2$ whenever $n = 42 k + 38.$

2    7   prime  7
8    6433   prime  7
14    4774777   prime  7
20    3486260113   prime  7



38    1350851580234038617   prime squared  49
80    147808829414345318853173402891795944513   prime squared  49
122    16173092699229880893715960009594875525837473033720099268457   prime squared  49

Your number is divisible by $121 = 11^2$ whenever $n = 110 k + 67.$

7    2123   prime  11
17    129074627   prime  11
27    7625530376123   prime  11



67    92709463147824050109467087204123   prime squared  121
177    2821383260958014531084804730393073172748132970923952481977527762896658545213494562627   prime squared  121
287    85861822891966292897565943276292392423908891501494514127947492094325821620603901184289283467528859878643948705742648123768724313989998123   prime squared  121

Your number is divisible by $289 = 17^2$ whenever $n = 272 k + 214.$

6    697   prime  17
22    31378962457   prime  17
38    1350851580234038617   prime  17



214    1270423474759653869629541561076150845942627861345583970679777076713806668073848513969400550694997546777   prime squared  289
486    7602033756829688179535612101927342434798006222913345882096671718462026450847558385638299237091029898106108915679642725019874527596206183615695170393375601813754958083630064304945006176793537681250116409274581709738622832815408017497   prime squared  289
758    45489491014727452017657094699993344217699387580459634625293727208970468768085650812024804178818092172750636489054819257623093543090018879398677204096366573883113560178809969964755425161715086488029060167818542608680433377185439106005853203184618487801892144052367301608312676367936439310746339829776474186019806821915640418802179770139744725661919759420980369817   prime squared  289

Surprising, your number is often divisible by $23$ but never by $529 = 23^2.$ Go Figure.

3    23   prime  23
14    4774777   prime  23
25    847271832227   prime  23
36    150094600937260753   prime  23
47    26588814288588759110123   prime  23
58    4710128697102129646845747817   prime  23

NO 23 SQUARED

Your number is divisible by $961 = 31^2$ whenever $n = 930 k + 828.$

18    387289417   prime  31
48    79766442936135021508033   prime  31
78    16423203268260507030504015972062417017   prime  31



828       prime squared  961
1758       prime squared  961
2688       prime squared  961
3618       prime squared  961
4548       prime squared  961

---

Your number is divisible by $1369 = 37^2$ whenever $n = 1332 k + 383.$

23    94138984523   prime  37
59    14130386091450504128613099323   prime  37
95    2120895147045314099684568958946760345244084523   prime  37



383       prime squared  1369
1715       prime squared  1369
3047       prime squared  1369
4379       prime squared  1369
5711       prime squared  1369

---

I also did a bounded factoring: given one of these numbers, use trial division with primes $p < 1200.$ I did catch a $1021^2$ this way..

jagy@phobeusjunior:~$ ./mse | grep "\^"
38     = 7^2 17  cdot mbox{BIG} 
67     = 11^2  cdot mbox{BIG} 
80     = 7^2 23 607  cdot mbox{BIG} 
122     = 7^2 137 599  cdot mbox{BIG} 
164     = 7^3 113  cdot mbox{BIG} 
177     = 11^2  cdot mbox{BIG} 
206     = 7^2 41  cdot mbox{BIG} 
214     = 17^2  cdot mbox{BIG} 
248     = 7^2  cdot mbox{BIG} 
287     = 11^2  cdot mbox{BIG} 
290     = 7^2 47 809 1033  cdot mbox{BIG} 
332     = 7^2 1193  cdot mbox{BIG} 
374     = 7^2 17 1087  cdot mbox{BIG} 
383     = 37^2  cdot mbox{BIG} 
397     = 11^3  cdot mbox{BIG} 
416     = 7^2 233  cdot mbox{BIG} 
458     = 7^3 439  cdot mbox{BIG} 
486     = 17^2 41  cdot mbox{BIG} 
500     = 7^2 113  cdot mbox{BIG} 
507     = 11^2 83  cdot mbox{BIG} 
508     = 73^2  cdot mbox{BIG} 
542     = 7^2 23  cdot mbox{BIG} 
584     = 7^2 431  cdot mbox{BIG} 
606     = 41^2  cdot mbox{BIG} 
617     = 11^2  cdot mbox{BIG} 
626     = 7^2  cdot mbox{BIG} 
668     = 7^2  cdot mbox{BIG} 
710     = 7^2 17 911  cdot mbox{BIG} 
712     = 271^2  cdot mbox{BIG} 
727     = 11^2 47  cdot mbox{BIG} 
752     = 7^3 89  cdot mbox{BIG} 
758     = 7 17^3  cdot mbox{BIG} 
794     = 7^2  cdot mbox{BIG} 
828     = 23 31^2 127 191  cdot mbox{BIG} 
836     = 7^2 113  cdot mbox{BIG} 
837     = 11^2 683  cdot mbox{BIG} 
878     = 7^2  cdot mbox{BIG} 
920     = 7^2  cdot mbox{BIG} 
947     = 11^2 983  cdot mbox{BIG} 
957     = 11 47^2 229  cdot mbox{BIG} 
962     = 7^2  cdot mbox{BIG} 
1004     = 7^2 23 937  cdot mbox{BIG} 
1030     = 17^2 151  cdot mbox{BIG} 
1046     = 7^3 17 41  cdot mbox{BIG} 
1057     = 11^2 59 431  cdot mbox{BIG} 
1088     = 7^2  cdot mbox{BIG} 
1130     = 7^2  cdot mbox{BIG} 
1167     = 11^2  cdot mbox{BIG} 
1172     = 7^2 113  cdot mbox{BIG} 
1214     = 7^2 569  cdot mbox{BIG} 
1256     = 7^2 47  cdot mbox{BIG} 
1277     = 11^2  cdot mbox{BIG} 
1298     = 7^2  cdot mbox{BIG} 
1302     = 17^2 47 223 263  cdot mbox{BIG} 
1340     = 7^3  cdot mbox{BIG} 
1382     = 7^2 17  cdot mbox{BIG} 
1387     = 11^2  cdot mbox{BIG} 
1424     = 7^2 479  cdot mbox{BIG} 
1466     = 7^2 23  cdot mbox{BIG} 
1491     = 83^2 157  cdot mbox{BIG} 
1497     = 11^2 433  cdot mbox{BIG} 
1508     = 7^2 113  cdot mbox{BIG} 
1550     = 7^2 727  cdot mbox{BIG} 
1574     = 7 17^2  cdot mbox{BIG} 
1588     = 73 191^2  cdot mbox{BIG} 
1592     = 7^2  cdot mbox{BIG} 
1607     = 11^3 37 167  cdot mbox{BIG} 
1634     = 7^5  cdot mbox{BIG} 
1676     = 7^2  cdot mbox{BIG} 
1715     = 37^2 587  cdot mbox{BIG} 
1717     = 11^2 1117  cdot mbox{BIG} 
1718     = 7^2 17  cdot mbox{BIG} 
1758     = 31^2  cdot mbox{BIG} 
1760     = 7^2  cdot mbox{BIG} 
1802     = 7^2  cdot mbox{BIG} 
1827     = 11^2  cdot mbox{BIG} 
1844     = 7^2 113 919  cdot mbox{BIG} 
1846     = 17^2 41  cdot mbox{BIG} 
1886     = 7^2 41 863  cdot mbox{BIG} 
1928     = 7^3 23  cdot mbox{BIG} 
1937     = 11^2  cdot mbox{BIG} 
1970     = 7^2  cdot mbox{BIG} 
2012     = 7^2  cdot mbox{BIG} 
2038     = 17 23 47^2  cdot mbox{BIG} 
2047     = 11^2  cdot mbox{BIG} 
2054     = 7^2 17  cdot mbox{BIG} 
2096     = 7^2  cdot mbox{BIG} 
2118     = 17^2 31  cdot mbox{BIG} 
2138     = 7^2  cdot mbox{BIG} 
2157     = 11^2  cdot mbox{BIG} 
2180     = 7^2 113  cdot mbox{BIG} 
2222     = 7^3 47  cdot mbox{BIG} 
2246     = 7 17 41^2  cdot mbox{BIG} 
2264     = 7^2  cdot mbox{BIG} 
2267     = 11^2  cdot mbox{BIG} 
2306     = 7^2 887  cdot mbox{BIG} 
2340     = 127^2  cdot mbox{BIG} 
2348     = 7^2 191  cdot mbox{BIG} 
2377     = 11^2 359  cdot mbox{BIG} 
2390     = 7^2 17^2 23 431  cdot mbox{BIG} 
2432     = 7^2  cdot mbox{BIG} 
2474     = 7^2  cdot mbox{BIG} 
2487     = 11^2 179  cdot mbox{BIG} 
2516     = 7^3 113  cdot mbox{BIG} 
2531     = 1021^2  cdot mbox{BIG} 
2558     = 7^2  cdot mbox{BIG} 
2597     = 11^2  cdot mbox{BIG} 
jagy@phobeusjunior:~$ 

---

---

Answer 3

The answer below has become out of date because the question changed. Please don't be that guy; ask a new question if you want to follow up.

We can show that there are multiples of $49$ using elementary modular arithmetic techniques.

Let $p$ be a prime greater than $3$ (why?), and seek values of $n$ for which

$$3^n\equiv 2^{n-1}\bmod p^2$$

Multiply by $(3^{-1})^{n-1}$ getting

$$3\equiv (2×3^{-1})^{n-1}\bmod p^2$$

Let us first try $p=5$. We die because the left side is a nonquadratic residue $\bmod p=5$ and the right side, with $2×3^{-1}\equiv 4\bmod 5$, is a quadratic residue.

Fortunately, for $p=7$ we avoid this contradiction because $2×3^{-1}$ is nonquadratic $\bmod 7$ thus also nonquadratic $\bmod 49$. We then have

$$3\equiv 17^{n-1}\bmod 49$$

where $17\equiv 3\bmod 7$ is a primitive root in the group of units $\bmod 49$ (the only nonprimitive root $\bmod 49$ congruent to $3\bmod 7$ is $31$), therefore this equation must have positive whole number solutions for $n$.

Will Jagy has identified the minimal solution as $n=38$, so let us check this case $\bmod 49$. Since units give $1$ when raised to the power of $42$, we may render

$$3^{38}\equiv (3^{-1})^4\equiv 33^4\equiv 11^2\equiv 121\equiv\color{blue}{23\bmod 49}$$

And

$$2^{37}\equiv (2^{-1})^5\equiv 25^5\equiv 25×(-12)^2\equiv 3600\equiv \color{blue}{23\bmod 49}$$