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I am working on this proof for a while now and can't think of any solution. Instinctively i think it has something to do with the formula

$x^n−y^n=(x-y)(x^{n-1}+x^{n-2}+\cdots+xy^{n−2}+y^{y−1})$

But i dont know how to go on with it. I suppose i can't use the method of induction?

I really appreciate your help!

Sil
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Jerry Cohen
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    This is not true. $29,|,(3^7-2^7)$. – lulu Oct 04 '20 at 12:45
  • ehmm 29 is a prime number – Jerry Cohen Oct 04 '20 at 12:48
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    The converse is true, namely "if $3^n-2^n$ is prime then $n$ is prime". Is that what you meant to ask? – lulu Oct 04 '20 at 12:48
  • Yes, $29$ is a prime number. So what? It's easy to see that $3^7-2^7\neq 29$ if that's what you were worried about. $3^7-2^7$ is quite a bit bigger than $29$. – lulu Oct 04 '20 at 12:48
  • $3^7-2^7 = 2059 = 29 \times 71$ is not prime. – Crostul Oct 04 '20 at 12:50
  • ah ok,, sorry i was confused. i do mean the converse. didnt realise that changing it would make the statement wrong. – Jerry Cohen Oct 04 '20 at 12:56
  • i now changed the question – Jerry Cohen Oct 04 '20 at 12:57
  • The edit is still wrong, and the same counterexample applies. – lulu Oct 04 '20 at 12:57
  • sorry, ive been working on proofs all morning, a bit brain dead right now.. – Jerry Cohen Oct 04 '20 at 13:01
  • Your formula establishes the converse quickly. Let's do it for, say, $3^{15}-2^{15}$. We can write that as $(3^3)^5-(2^3)^5$ which we know (by what you wrote) is divisible by $3^3-2^3=19$ Indeed, $3^{15}-2^{15}=19\times 211\times 3571$. Generalize this. A quick corollary to this approach is that, if $n$ is even, $3^n-2^n$ is divisible by $3^2-2^2=5$. – lulu Oct 04 '20 at 13:08
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    Hint: Assume $n$ is not prime, then $n=n_1 \cdot n_2$ where both $n_1 , n_2 >1$. – rtybase Oct 04 '20 at 13:17
  • if $c > 13$ is a fixed positive integer, then the exponential Diophantine equation

    $$|3^x − 2^y| = c$$

    admits at most one solution in positive integers $x$ and $y$.

    R. J. Stroeker and R. Tijdeman, Diophantine equations, in Computational methods in number theory, Part II, vol. 155 of Math. Centre Tracts, Math. Centrum, Amsterdam, 1982, pp. 321–369.

    Also, see: https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/PerfectPowers.pdf

     – vvg Oct 04 '20 at 13:21

1 Answers1

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Suppose , $\ n\ $ is composite and $\ p\ $ a prime factor of $\ n\ $

Then , we have $3^p-2^p\mid 3^n-2^n$ which can be seen by setting $x:=3^p$ , $y:=2^p$ and considering $x-y\mid x^k-y^k$ for every positive integer $k$.

To show that $3^p-2^p$ is a proper factor, it is sufficient to prove $1<3^p-2^p<3^n-2^n$

$$f(x)=3^x-2^x$$ has derivate $$f'(x)=3^x\ln(3)-2^x\ln(2)$$ which is positive for $x\ge 0$ because of $3^x\ge2^x>0$ and $\ln(3)>\ln(2)>0$ , hence $f(x)$ is strictly increasing for $x\ge 0$

Therefore because of $1<p<n$ we have $1<3^p-2^p<3^n-2^n$

Since $n=1$ gives $1$ which is not prime, a prime can only occur if $n$ itself is prime.

Peter
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  • Whenever I see maths like this it increases my conviction that the Collatz conjecture is connected to some deep foundational maths. – Robert Frost Dec 01 '20 at 17:06