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Motivation :

I wanted to understand more the behaviour of the divisors of $3^n-2^n$,

to solve this conjecture : Conjecture : $\forall n>1, d\mid 3^n-2^n \Rightarrow v_2(d+1)<n$

So I tried to examine initially: $\tau(3^n-2^n)$ (the number of divisors $d$ of $3^n-2^n$)

My attempt:

I literally have no idea of how we can prove this result,

but I discover it from my python research routine !

from sympy import divisor_count from sympy.ntheory import factorint for n in range(2,100): print(n,factorint(divisor_count(pow(3,n)-pow(2,n))))

Here's a chart of the factorization of $\tau(3^n-2^n)$ , ($\forall n < 10^2 $)

$\color{blue}{2}[\color{green}{2}\color{red}:1]$ $\color{blue}{3}[\color{green}{2}\color{red}:1]$ $\color{blue}{4}[\color{green}{2}\color{red}:2]$ $\color{blue}{5}[\color{green}{2}\color{red}:1]$ $\color{blue}{6}[\color{green}{2}\color{red}:3]$ $\color{blue}{7}[\color{green}{2}\color{red}:2]$ $\color{blue}{8}[\color{green}{2}\color{red}:3]$ $\color{blue}{9}[\color{green}{2}\color{red}:2]$ $\color{blue}{10}[\color{green}{2}\color{red}:2,\color{green}{3}\color{red}:1]$ $\color{blue}{11}[\color{green}{2}\color{red}:1,\color{green}{3}\color{red}:1]$ $\color{blue}{12}[\color{green}{2}\color{red}:5]$ $\color{blue}{13}[\color{green}{2}\color{red}:2]$ $\color{blue}{14}[\color{green}{2}\color{red}:4]$ $\color{blue}{15}[\color{green}{2}\color{red}:3]$ $\color{blue}{16}[\color{green}{2}\color{red}:5]$ $\color{blue}{17}[\color{green}{2}\color{red}:1]$ $\color{blue}{18}[\color{green}{2}\color{red}:5]$ $\color{blue}{19}[\color{green}{2}\color{red}:2]$ $\color{blue}{20}[\color{green}{2}\color{red}:4,\color{green}{3}\color{red}:1]$ $\color{blue}{21}[\color{green}{2}\color{red}:5]$ $\color{blue}{22}[\color{green}{2}\color{red}:3,\color{green}{3}\color{red}:1]$ $\color{blue}{23}[\color{green}{2}\color{red}:2]$ $\color{blue}{24}[\color{green}{2}\color{red}:7]$ $\color{blue}{25}[\color{green}{2}\color{red}:3]$ $\color{blue}{26}[\color{green}{2}\color{red}:5]$ $\color{blue}{27}[\color{green}{2}\color{red}:3]$ $\color{blue}{28}[\color{green}{2}\color{red}:6]$ $\color{blue}{29}[\color{green}{2}\color{red}:1]$ $\color{blue}{30}[\color{green}{2}\color{red}:7,\color{green}{3}\color{red}:1]$ $\color{blue}{31}[\color{green}{2}\color{red}:1]$ $\color{blue}{32}[\color{green}{2}\color{red}:7]$ $\color{blue}{33}[\color{green}{2}\color{red}:4,\color{green}{3}\color{red}:1]$ $\color{blue}{34}[\color{green}{2}\color{red}:4]$ $\color{blue}{35}[\color{green}{2}\color{red}:5]$ $\color{blue}{36}[\color{green}{2}\color{red}:10]$ $\color{blue}{37}[\color{green}{2}\color{red}:2]$ $\color{blue}{38}[\color{green}{2}\color{red}:5]$ $\color{blue}{39}[\color{green}{2}\color{red}:4]$ $\color{blue}{40}[\color{green}{2}\color{red}:8,\color{green}{3}\color{red}:1]$ $\color{blue}{41}[\color{green}{2}\color{red}:4]$ $\color{blue}{42}[\color{green}{2}\color{red}:8,\color{green}{3}\color{red}:1]$ $\color{blue}{43}[\color{green}{2}\color{red}:3]$ $\color{blue}{44}[\color{green}{2}\color{red}:5,\color{green}{3}\color{red}:1]$ $\color{blue}{45}[\color{green}{2}\color{red}:5]$ $\color{blue}{46}[\color{green}{2}\color{red}:6]$ $\color{blue}{47}[\color{green}{2}\color{red}:3]$ $\color{blue}{48}[\color{green}{2}\color{red}:10]$ $\color{blue}{49}[\color{green}{2}\color{red}:4]$ $\color{blue}{50}[\color{green}{2}\color{red}:8]$ $\color{blue}{51}[\color{green}{2}\color{red}:5]$ $\color{blue}{52}[\color{green}{2}\color{red}:7,\color{green}{3}\color{red}:1]$ $\color{blue}{53}[\color{green}{2}\color{red}:1]$ $\color{blue}{54}[\color{green}{2}\color{red}:7]$ $\color{blue}{55}[\color{green}{2}\color{red}:4,\color{green}{3}\color{red}:1]$ $\color{blue}{56}[\color{green}{2}\color{red}:9]$ $\color{blue}{57}[\color{green}{2}\color{red}:4,\color{green}{3}\color{red}:1]$ $\color{blue}{58}[\color{green}{2}\color{red}:5]$ $\color{blue}{59}[\color{green}{2}\color{red}:1]$ $\color{blue}{60}[\color{green}{2}\color{red}:12,\color{green}{3}\color{red}:1]$ $\color{blue}{61}[\color{green}{2}\color{red}:3]$ $\color{blue}{62}[\color{green}{2}\color{red}:5]$ $\color{blue}{63}[\color{green}{2}\color{red}:9]$ $\color{blue}{64}[\color{green}{2}\color{red}:9]$ $\color{blue}{65}[\color{green}{2}\color{red}:6]$ $\color{blue}{66}[\color{green}{2}\color{red}:9,\color{green}{3}\color{red}:1]$ $\color{blue}{67}[\color{green}{2}\color{red}:3]$ $\color{blue}{68}[\color{green}{2}\color{red}:6]$ $\color{blue}{69}[\color{green}{2}\color{red}:8]$ $\color{blue}{70}[\color{green}{2}\color{red}:8,\color{green}{3}\color{red}:1]$ $\color{blue}{71}[\color{green}{2}\color{red}:2]$ $\color{blue}{72}[\color{green}{2}\color{red}:15]$ $\color{blue}{73}[\color{green}{2}\color{red}:3]$ $\color{blue}{74}[\color{green}{2}\color{red}:5]$ $\color{blue}{75}[\color{green}{2}\color{red}:8]$ $\color{blue}{76}[\color{green}{2}\color{red}:7]$ $\color{blue}{77}[\color{green}{2}\color{red}:6,\color{green}{3}\color{red}:1]$ $\color{blue}{78}[\color{green}{2}\color{red}:10]$ $\color{blue}{79}[\color{green}{2}\color{red}:4]$ $\color{blue}{80}[\color{green}{2}\color{red}:11,\color{green}{3}\color{red}:1]$ $\color{blue}{81}[\color{green}{2}\color{red}:6]$ $\color{blue}{82}[\color{green}{2}\color{red}:9]$ $\color{blue}{83}[\color{green}{2}\color{red}:3]$ $\color{blue}{84}[\color{green}{2}\color{red}:13,\color{green}{3}\color{red}:1]$ $\color{blue}{85}[\color{green}{2}\color{red}:4]$ $\color{blue}{86}[\color{green}{2}\color{red}:9]$ $\color{blue}{87}[\color{green}{2}\color{red}:5]$ $\color{blue}{88}[\color{green}{2}\color{red}:9,\color{green}{3}\color{red}:1]$ $\color{blue}{89}[\color{green}{2}\color{red}:2]$ $\color{blue}{90}[\color{green}{2}\color{red}:12,\color{green}{3}\color{red}:1]$ $\color{blue}{91}[\color{green}{2}\color{red}:8]$ $\color{blue}{92}[\color{green}{2}\color{red}:9]$ $\color{blue}{93}[\color{green}{2}\color{red}:4]$ $\color{blue}{94}[\color{green}{2}\color{red}:7]$ $\color{blue}{95}[\color{green}{2}\color{red}:6]$ $\color{blue}{96}[\color{green}{2}\color{red}:15]$ $\color{blue}{97}[\color{green}{2}\color{red}:2]$ $\color{blue}{98}[\color{green}{2}\color{red}:9]$ $\color{blue}{99}[\color{green}{2}\color{red}:7,\color{green}{3}\color{red}:1]$

Thanks for help !

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  • $\begingroup$ Also , I wonder if someone could find a closed formula of $a,b$ where $\tau(3^n-2^n)=2^a\cdot 3^b$ $\endgroup$ Commented Jun 29 at 18:07
  • $\begingroup$ Your search isn't nearly far enough. After all, to find a small number $n$ for which $\tau(n)$ is not of that form, you'd first look for a prime $p$ such that $v_p(n)=4$. You won't find that with small numbers. $\endgroup$ Commented Jun 29 at 18:10
  • $\begingroup$ did you meant $\tau(3^n-2^n)$ or $\tau(n)$ ? $\endgroup$ Commented Jun 29 at 18:11
  • $\begingroup$ I meant $\tau(n)$. After all, your claim is trivially true for most small natural numbers. $\endgroup$ Commented Jun 29 at 18:12

2 Answers 2

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Your search only involves quite small numbers. Counterexamples would need something like a prime $p$ such that $v_p(3^n-2^n)=4$ or $=6$, something like that. Since clearly $p>3$, you can't do this with small numbers.

That said, we can construct a counterexample. We can solve $$3^n\equiv 2^n\pmod {5^4}\iff n\equiv 0,\,250 \pmod {500}$$

And it is easy to check that $3^{250}\not \equiv 2^{250}\pmod {5^5}$, so $5\,|\,\tau \left(3^{250}-2^{250}\right)$. Thus $n=250$ is a counterexample.

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  • $\begingroup$ Thanks a lot ! I've understood now that small numbers can't tell you a lot about divisibility properties ! $\endgroup$ Commented Jun 29 at 18:22
  • $\begingroup$ I have a question though, can we relate $\tau(3^n-2^n)$ with $\tau(n)$ like LTE does between $v_p(3^n-2^n)$ and $v_p(n)$ ? $\endgroup$ Commented Jun 29 at 18:22
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    $\begingroup$ @Lhachimi I wouldn't think so. Easy enough to compute for small $n$ to see if any pattern appears to hold. $\endgroup$ Commented Jun 29 at 18:26
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$3^{250}-2^{250}=5^4\cdot11^1\cdot101^1\cdot211^1\cdot251^1\cdot1201^1\cdot513101^1\cdot39756701^1\cdot27884532843001^1$

$\cdot14726555364585012892444440090751^1 \cdot515397931586890805724262529672368870694338434001^1$

$\Rightarrow \tau(3^{250}-2^{250})=5\cdot2^{10} \not= 2^a \cdot 3^b$

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  • $\begingroup$ The factorization took ~$7$ minutes (using PARI/GP) $\endgroup$ Commented Jul 1 at 8:31
  • 1
    $\begingroup$ You can get that factorization almost instantly if you ask it to factor the obvious factors $3^{125}-2^{125}$ and $3^{125}+2^{125}$ separately. $\endgroup$ Commented Jul 9 at 18:36
  • $\begingroup$ Yeah, it's a wonderful idea tnx :) $\endgroup$ Commented Jul 9 at 18:59
  • $\begingroup$ $$ \begin{aligned} \text{factor}(3^{125} - 2^{125}) &= 101 \cdot 211 \cdot 39756701 \cdot 515397931586890805724262529672368870694338434001 \\ \\ \text{factor}(3^{125} + 2^{125}) &= 5^4 \cdot 11 \cdot 251 \cdot 1201 \cdot 513101 \cdot 27884532843001 \cdot 14726555364585012892444440090751 \end{aligned} $$ $\endgroup$ Commented Jul 9 at 19:00

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