Intereting Posts

Show: $\mathbb{E}(f|\mathcal{F})=\mathbb{E}(f)$
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A subsequence of a convergent sequence converges to the same limit. Questions on proof. (Abbott p 57 2.5.1)
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proving an invloved combinatorial identity
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Abstract algebra: for a polynomial $p$, prove $\sigma(\tau(p))=(\sigma\tau)(p)$ for all $\sigma, \tau \in S_n$
Gradients of marginal likelihood of Gaussian Process with squared exponential covariance, for learning hyper-parameters
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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:

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

- $y^2 = \frac{x^5 - 1}{x-1}$ & $x,y \in \mathbb{Z}$
- Exponent of $p$ in the prime factorization of $n!$
- Proving that an integer is even if and only if it is not odd
- Using the Euler totient function for a large number
- What is the Euler Totient of Zero?
- Integral solutions to $1\times2+2\times3+\cdots+m\times(m+1)=n\times(n+1)$.

But that gets messy…

- Divisibility and the Fibonacci sequence
- Prove that $\sqrt 5$ is irrational
- Computing the Legendre symbol $\bigl(\frac{3}{p}\bigr)$ using Gauss' Lemma
- If all of the integers from $1$ to $99999$ are written down in a list, how many zeros will have been used?
- Integer $2 \times 2$ matrices such that $A^n = I$
- Divisibility by $9$
- Roman Numbers - Conversion to decimal number
- If $p$ is prime, then $n\mid\varphi(p^n-1)$
- Number of solutions of $x^2=1$ in $\mathbb{Z}/n\mathbb{Z}$
- 2011 AIME Problem 12, probability round table

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.

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