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Trying to prove that $(3^n - 2^n)/n$ is not an integer for $n\geq 2$.

Was trying something along the lines of induction with:

$3^{n+1} - 2^{n+1} = 2(3^n - 2^n) + 3^n \equiv 0 \mod (n+1)$

But that gets messy...

t.b.
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user10084
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3 Answers3

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Let $p$ be the smallest prime factor of $n$ and write $n = p^k m$ where $p \nmid m$. By repeated application of Fermat's little theorem it follows that $3^n - 2^n \equiv 3^m - 2^m \equiv 0 \bmod p$. If $p = 2, 3$ then this is clearly never possible; otherwise, it follows that $\left( \frac{2}{3} \right)^m \equiv 1 \bmod p$, hence that $\gcd(p-1, m) > 1$. But this is also impossible since the prime factors of $m$ are all larger than $p$.

Qiaochu Yuan
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    Note: Being a homogenization of this prior problem, the proof is just a homogenization of its proof. – Bill Dubuque Apr 26 '11 at 19:12
  • @Bill, What does that mean? Homogenization? – quanta Apr 26 '11 at 19:21
  • Great solution! How did you know to look at the smallest prime factor? – quanta Apr 26 '11 at 19:22
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    The homogenization of a polynomial $\rm:f(x)\ =\ a_n x^n + \cdots + a_1\ x + a_0\ $ is the homogeneous polynomial $\rm:y^n\ f(x/y)\ =\ a_n x^n + \cdots + a_1\ x\ y^{n-1} + a_0\ y^n:.:$ In particular, $\rm: x^n - 1\ \to\ x^n - y^n:,:$ explaining the intimate connection between the two problems. – Bill Dubuque Apr 26 '11 at 19:31
  • edit: nevermind I see it now – user10084 Apr 26 '11 at 19:35
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    @quanta: to be honest, I've seen this problem before. It's an old chestnut, as they say. I guess the idea is to start with the case that $n$ is prime and then see how far you can generalize that argument, picking any prime factor of $n$ at first, and then noting that you get a contradiction if the factor you chose is minimal. – Qiaochu Yuan Apr 26 '11 at 19:38
  • @Qia As I mentioned in the prior problem, one way to view the essence of the problem is by dehomogenizing and canonicalizing, reducing to the fact that if $\rm\ A^N = 1\ $ then the order of $\rm:A:$ is $\ge$ the least prime factor $\rm:P:$ of $\rm:N:$. In particular this implies that $\rm\ A^{P-1}\ne 1:,\ $ which settles the problem at hand. – Bill Dubuque Apr 26 '11 at 20:57
  • @QiaochuYuan I couldn't understand your 2nd statement "By repeated application of Fermat's little theorem it follows that......".Can you insert some more steps so that I can understand. – Saurabh Jun 10 '12 at 19:04
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    @Saurabh: repeated applications of Fermat's little theorem show that if $p-1 | n - m$ and $n, m \ge 1$ then $a^n \equiv a^m \bmod p$. Now let $n = p^k m$ as above. We have $n - m = (p^k - 1) m$ which is divisible by $p-1$. – Qiaochu Yuan Jun 10 '12 at 19:46
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If $$ (3^n−2^n)/n $$ is an integer, then $$n |(3^n−2^n)$$ i.e., $$3^n−2^n=kn $$ for some integer k. $$ 3^n=2^n+kn. $$ If kn is even, then the LHS must be even, so $$2∤n.$$ $$ 2^n=3^n-kn $$ If 3|kn, then 3 will divide $$2^n$$(the LHS), so $$3∤n.$$ It implies n is co-prime to 3*2 i.e, (3*2,n)=1. $$3^n−2^n=kn =>3^n≡2^n(mod\ n)=>(3*2^{-1})^n≡1(mod\ n)$$

If p is the smallest prime that divides n, then $$(3*2^{-1})^n≡1(mod\ p)$$ If $$ ord_p(3*2^{-1})=D,\ then\ D|(p-1,n).$$ But. p being the smallest factor of n, n can not have any factor>1 common with p-1 =>D=1=>3≡2(mod p) i.e., 1|p.

Observation: If $$ ord_n(3*2^{-1})=d,$$ then the problem reduces to find d such that d|n and d|φ(n).

lab bhattacharjee
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I would consider first the case that $n \geq 3$ is prime. Using Fermat's Little Theorem, you should be able to show $3^{n} - 2^{n} \equiv 1 \text{ (mod }n)$.

EDIT: I retract what I said about the composite case. I'll keep thinking about it.

Hans Parshall
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  • @Hans: "Argue this is not the case by the induction hypothesis": I don't get it. The induction hypothesis says that $3^p-2^p$ is not divisible by $p$; why does this imply $3^{n+1}-2^{n+1}$ is not divisible by $p$? Perhaps I'm being stupid... – TonyK Apr 26 '11 at 16:08
  • Suppose pk = n+1. Then 3^(n+1) = 3^(pk) = (3^k)^p and now you can use Fermat's little theorem. – Charles Apr 26 '11 at 17:05
  • Equivalently, you can use the lemma that $p|n$ implies that $(a^p-b^p)|(a^n-b^n)$ (which can be shown via an easy explicit factorization - with $n=pq$, say, let $c=a^p$, $b=d^p$; then the lemma is equivalent to $(c-d)|(c^q-d^q)$ ) – Steven Stadnicki Apr 26 '11 at 17:09
  • You don't need any induction Charle's remark above helps use fermat when n is not prime. moreover you need to notice that any power of 3 and n are coprime same goes for any power of 2 and n. At the end one could generalize the question by putting a instead of 3 and b instead of 2 with the condition that a and b are coprime. – El Moro Apr 26 '11 at 17:20
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    Comments are appreciated but I still don't see how this solves the problem. If anyone sees it clearly, maybe they should post an answer. – Hans Parshall Apr 26 '11 at 17:51
  • Charles, not seeing how I could use Fermat's little theorem since the modulus would be still be composite. – user10084 Apr 26 '11 at 18:20