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Having been interested by the question Solution to $|2^m-3^n|=35$? I was making some computations and I observed that numbers of the form $3^{12k-2i}- 2^{12k-2j}$, where $k$ any natural number, $i = \{0,1,2,\dots 6k\}$ and $ j=6k-i$,

are always divisible by $35$.

For example:
if $k=1, i=1, j=5$ then $3^{10} - 2^2= 5\cdot 7^2\cdot241$
if $k=2, i=3 , j= 9$ then $3^{18}-2^6 = 5^2\cdot 7\cdot 29\cdot 97 \cdot 787$, etc..

Moreover it seems that this formula covers all numbers of the form $3^n-2^m$ which are divisible by $35$.

  • How to prove the case ?

Additionally $2,3,5,7$ are consecutive primes.

  • Are there formulas for other consecutive primes when $p_1^m-p_2^n$ is divisible by $p_3 \cdot p_4$ ? How to arrive at them by other method than experimental calculations?
Widawensen
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  • A provisorical (and not complete but quick) observation is, that $(3^n-1) - (2^m-1)$ is divisible by $35$ if each of its parentheses are divisible by $35$. But for the parentheses to be divisible by $35$ there are relatively simple formulae depending on $n$ and $m$. ($n$ and $m$ must be multiples of $12$) Of course this gives only a subset of possible solutions... – Gottfried Helms Nov 08 '17 at 10:42
  • @GottfriedHelms Yes, I was aware of that.. I wanted to have formula for all possible multiplies of $35$ for $3^n-2^m$ and probably the listed one in the question is the searched formula but this is supported only by experimental evidence so far.. – Widawensen Nov 08 '17 at 10:57
  • Whatever the factorization would be always we have in it $5$ and $7 \ \ \ $ $2^{142}-3^2 = 571714488917196741949329472236648286964521369 $ – Widawensen Nov 08 '17 at 11:20
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    What I find by q&d observation is: let $n=2N$ and $m=2M$ then if $n+m \equiv 0 \pmod {12}$ the expression is divisble by $35$. Surely this has to do with the cyclic group-order of $5 $ modulo $2$ and modulo$ 3$ which are both $4$ and that of $7$ which is $3$ and $6$ respectively. This leads to the constant modulus $\text{lcm}(4,6)=12$ in that observation... – Gottfried Helms Nov 08 '17 at 11:21
  • @GottfriedHelms Very important observation..maybe even generalization is possible taking this into account.. – Widawensen Nov 08 '17 at 11:25
  • The nonnegative integers $(n,m)$ such that $3^n-2^m$ is a multiple of $35$ are exactly the couples $$(n,m)=(2i-12k,12\ell-2i)$$ with $(i,k,\ell)$ integers such that $$6k\leqslant i\leqslant6\ell$$ (The condition that your question asserts is not equivalent to this.) – Did Nov 08 '17 at 11:58
  • @Widawensen: don't mind. If I'm online then I'll see how you proceed(ed) – Gottfried Helms Nov 08 '17 at 12:10
  • We know that if $a ł x$ and $a ł$ y then $a| x^k -y^k$ , if $a| x-y$ .

    Consider $3^6-2^6=729-64=665=35 . 19$ ⇒ $(3^6)^k -(2^6)^k≡ 0 mod 35$ or:

    $ 3 ^{6k} -2^{6k} ≡ 0 mod 35$

    The difference of two consecutive prime (or any two primes except 2 and 3)is an even number which is composite and can be factorized into primes which are less than $p_1$ and $ p_2$, that is there can be more than two primes in factors(like in you 2nd example).

     – sirous Nov 08 '17 at 20:10
  • @Did Hmm, I wonder what a pair $(n,m)$ is generated with your formula but not generated with mine ... could you give the example? – Widawensen Nov 09 '17 at 09:49
  • @Widawensen Hmm, it seems the two representations actually yield the same couples $(n,m)$. Maybe the one in your post can be somewhat streamlined to $$(n,m)=(12k-2i,2i)\qquad0\leqslant i\leqslant6k$$ – Did Nov 09 '17 at 17:19
  • @Did Thank you for the confirmation of equivalence. Really the more transformable form is presented by you and Gottfried in his answer, maybe I unnecessary wanted to make mine more symmetrical ( then it is corresponding to the symmetry in visual presentation of numbers in the table) , but easier for transformation is however your form. – Widawensen Nov 10 '17 at 09:49
  • @Widawensen No problem (and apologies for my first reading, obviously too hasty). – Did Nov 10 '17 at 10:05

1 Answers1

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A solution is $$3^{2m'}-2^{12n'-2m'} \equiv 0 \pmod{35}\qquad \qquad \text{or} \\9^{m'}-4^{6n'-m'} \equiv 0 \pmod{35} \qquad \qquad \phantom {\text{or} } $$ Perhaps this suffices for a formal proof (but I do not see at the moment that the resulting equality is necessary AND ALSO sufficient: $$(4\cdot 9)^{m'}-4^{6n'} \equiv 0 \pmod{35} \\ 36^{m'}-4^{6n'} \equiv 0 \pmod{35} \\ 1-(2^{4 \cdot 3})^{ n'} \equiv 0 \pmod{35} \\ 1-1^{ n'} \equiv 0 \pmod{35} \\ 1-1 \equiv 0 \pmod{35} \\ $$

The rest of the proof needs to introduce the cyclic multiplicative order of $5$ and $7$ for the base $2$ (which is $4$ and $3$ respectively and gives $\text{lcm}(4,3)=12$ for $5 \cdot 7$), however to nicely write out the last steps it should be easy enough to leave it to you...

Gottfried Helms
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  • Thank you Gottfried for the solution, it's really hard to say whether it gives all conditions, I have to consider it .. – Widawensen Nov 09 '17 at 10:01
  • The formula is proven by Gottfried very smoothly but how to be sure that formula covers all possible numbers $3^m - 2^n $ divisible by $35$? – Widawensen Nov 09 '17 at 10:25
  • If $q≠p$ , ( in this case $5$ and $7$ ) then CarmichaelFunction $λ(pq)=lcm(λ(p),λ(q))=lcm(p−1,q−1)$ Nice formula for an expression of two numbers (in this case $2$ and $3$ : $3^n−2^m$) is supported by the condition that they have the same $λ(pq)$ what is also the value for other primes coprime with $p$ and $q$. – Widawensen Nov 09 '17 at 16:12
  • Interesting situation is also for $11^n-2^m$. Let $p=3,q=7$ then we have $\lambda(3\cdot 7)= 6$ and $2^2\cdot 11^2 \equiv 1 \pmod{ 21}$ – Widawensen Nov 09 '17 at 16:26
  • For $19^n−2^m$. Let $p=3,q=13$ then we have $ λ(3⋅13)=12$ and $2^2⋅19^2≡1 \pmod {39})$ – Widawensen Nov 09 '17 at 16:40
  • However the most interesting pattern is for $17^m-2^n$, here $2^2 \cdot 17^2 \equiv {1} \pmod {3 \cdot 5} $ but also $\pmod {3 \cdot 7}$ and $\pmod {5 \cdot 7}$ – Widawensen Nov 09 '17 at 16:52
  • @Widawensen: won't you like to make an collecting/concluding answer from your observations? – Gottfried Helms Nov 09 '17 at 16:59
  • Yes, I'm moving toward the searched generalization unfortunately I have to leave now MSE till tomorrow, so if you would like to add it to your answer Gottfried feel free to do it.. generalization could be based on your proof it seems.. – Widawensen Nov 09 '17 at 17:03