Prove that $3^n|4^{3^{n-1}}-1$, but $3^{n+1}\not|4^{3^{n-1}}-1$ for all $n\in\mathbb N$.

Question

Prove that $3^n|4^{3^{n-1}}-1$, but $3^{n+1}\not|4^{3^{n-1}}-1$ for all $n\in\mathbb N$.

I am unfortunately stuck in trying to prove this problem. It is obviously an exercise in mathematical induction. It is extremely easy to show the base cases for mathematical induction for $n=1$ and $n=2$, so I will not do those here.

I assume that our inductive hypothesis is to say that $3^k|4^{3^{k-1}}-1$, but $3^{k+1}\not|4^{3^{k-1}}-1$ for some natural number $k$. Now we need to show that $3^{k+1}|4^{3^{k}}-1$, but $3^{k+2}\not|4^{3^{k}}-1$. I started off by letting $x^3 = 4^{3^{k}}$. Then $x = \left(4^{3^{k}}\right)^{1/3} = 4^{3^{k}\over 3} = 4^{3^{k-1}}$. So, $$\begin{align}4^{3^k} - 1 &= x^3 - 1\\&=(x-1)(x^2+x+1) \tag{Difference of cubes}\\ &=(4^{3^{k-1}}-1)(4^{2\cdot3^{k-1}}+4^{3^{k-1}}+1).\end{align}$$ Something isn't quite lining up with what I expect to happen. We know that $3^{k+1}$ is not supposed to divide $4^{3^{k-1}}-1$ by our inductive hypothesis. So, this won't cancel out in the above like I'd hope for it to. That would have to mean the other stuff will cancel out, but it is not immediately clear how this happens. Or did I miss something completely important?

For the record, a much younger student brought this question to me, and I just got a little stuck in trying to work it out.

Answer 1

While you have approached this with mathematical induction, there is a general theorem for this kind of problem.

Let us use the LTE Lemma. The LTE Lemma states that for any odd prime $p$ and for any $x,y$ that are not divisible by $p$, we have that $$\nu_{p}(x^{n}-y^{n})=\nu_{p}(x-y)+\nu_{p}(n) $$Where $\nu_{p}(n)$ denotes the largest power of $p$ that divides $n$.Note that we have that for $p=3,x=4,y=1$, $$\nu_{3}\left(4^{3^{n-1}}-1\right)=\nu_{3}(4-1)+\nu_{3}(3^{n-1})=1+n-1=n$$ So $3^{n}$ is the largest power of $3$ that is a divisor of $4^{3^{n-1}}-1$. So our proof is done.

Of course we could use induction.

We are examining $4^{2\cdot3^{k-1}}+4^{3^{k-1}}+1$, but first let us examine $4^{3^{k-1}}=(3+1)^{3^{k-1}}=x$.

By the binomial theorem, we get that $x$ is of the form $3m+1$ where $m$ is an integer. Putting this back into our original equation, note that $$4^{2\cdot3^{k-1}}+4^{3^{k-1}}+1=(3m+1)^2+(3m+1)+1=9m^2+9m+3 \equiv 3 \pmod {9} $$So it is not divisible by $9$ but it is divisible by $3$. So our proof is done.

Answer 2

You wrote

$$\begin{align}4^{3^k} - 1 &= x^3 - 1\\&=(x-1)(x^2+x+1) \tag{Difference of cubes}\\ &=(4^{3^{k-1}}-1)(4^{2\cdot3^{k-1}}+4^{3^{k-1}}+1).\end{align}$$

Note here that $$4^{2\cdot 3^{k-1}}+4^{3^{k-1}}+1\equiv 1^{2\cdot 3^{k-1}}+1^{3^{k-1}}+1\equiv 1+1+1\equiv 0\pmod 3$$ and that for $k\ge 2$, using $4^3\equiv 1\pmod 9$, $$4^{2\cdot 3^{k-1}}+4^{3^{k-1}}+1\equiv (4^3)^{2\cdot 3^{k-2}}+(4^3)^{3^{k-2}}+1\equiv 1+1+1\equiv 3\not\equiv 0\pmod{9}$$ So, the induction works.