# Proof that $(3^n – 2^n) / n$ is not an integer, $n \geq 2$

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…

#### Solutions Collecting From Web of "Proof that $(3^n – 2^n) / n$ is not an integer, $n \geq 2$"

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 $p$.

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).

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.