Prove that $\frac{a^n-1}{b^n-1}$ and $\frac{a^{n+1}-1}{b^{n+1}-1}$ can't both be prime.

Prove that $$\frac{a^n-1}{b^n-1} \ \text{and} \ \frac{a^{n+1}-1}{b^{n+1}-1}$$ cannot both be prime ($a>b>1,n\ge 2$).

Clearly $(a^n-1,a^{n+1}-1)=a-1$ and $(b^n-1,b^{n+1}-1)=b-1$.

Clearly $b-1\mid a-1$ (since $b-1\mid a^n-1,a^{n+1}-1$), so $a=bk-k+1$ for some $k\in\mathbb Z_{\ge 2}$.

A hint would probably be enough. I haven't found an approach I could use.

Solution 1:

Let $\let\v\nu\let\geq\geqslant\v_p(n)$ denote the exponent of the prime $p$ in the factorisation of $n\in\mathbb N$.

Lemma (Lifting The Exponent) If $p$ is odd, $p\nmid a$ and $p\mid a-b$, then $$\v_p(a^n-b^n)=\v_p(a-b)+\v_p(n).$$ Proof. See ProofWiki.

We already have $b-1\mid a-1$. Let $r$ be any prime divisor of $b-1$ and thus of $a-1$. If $r$ is odd, then (by the lemma) $$\v_r\left(\frac{a^n-1}{b^n-1}\right)=\v_r(a-1)-\v_r(b-1).$$ Because $\frac{a^n-1}{b^n-1}=p$ is prime, this shows that $\v_r(a-1)=\v_r(b-1)$ for all primes $r\mid b-1$ with $r\neq2$ and $r\neq p$. Hence $a-1=2^kp^xg(b-1)$ for some $k,x$ and $g$, $\gcd(g,b-1)=1$.

We can do the same for $\frac{a^{n+1}-1}{b^{n+1}-1}=q$, giving $a-1=2^lq^yh(b-1)$ for some $l,y$ and $h$, $\gcd(h,b-1)=1$.

Note that $a-1=2^lq^yh(b-1)$ uniquely determines $q$ if $q\mid b-1$ and $\gcd(n,b-1)=1$:

• If $q$ is odd, it is the unique odd prime $r\mid b-1$ with $\v_r(a-1)>\v_r(b-1)$
• $q=2$ iff there is no odd prime $r\mid b-1$ with $\v_r(a-1)>\v_r(b-1)$

So we actually have that $p=q$: $$\frac{a^{n+1}-1}{b^{n+1}-1}=\frac{a^n-1}{b^n-1}$$ hence $$a^{n+1}b^n-a^{n+1}-b^n=b^{n+1}a^n-b^{n+1}-a^n$$ $$\color{darkgreen}{(a-b)}\color{blue}{(a^nb^n+\frac{a^n-b^n}{a-b}-\frac{a^{n+1}-b^{n+1}}{a-b})}=0$$ but, because $b^n\geq2^n\geq n+1$, $$\color{blue}{a^nb^n+\frac{a^n-b^n}{a-b}}\geq (n+1)a^n\color{blue}>a^n+a^{n-1}b+\cdots+ab^{n-1}+b^n=\color{blue}{\frac{a^{n+1}-b^{n+1}}{a-b}}$$ contradicting $\color{darkgreen}{a\neq b}$.

Solution 2:

I work out an incomplete proof, maybe you can complete it.

Suppose that both numbers are primes, that is :

$$\frac{a^n-1}{b^n-1}=p_1 \:\:\:\:\text{and} \:\:\:\:\ \frac{a^{n+1}-1}{b^{n+1}-1}=p_2$$

Now I can rearrange and factorize $a^n-1$ and $a^{n+1}-1$ in this way:

$$(a-1)(a^{n-1}+a^{n-2}+...+a+1)=p_1 (b^n-1)$$ $$(a-1)(a^{n}+a^{n-1}+...+a+1)=p_2 (b^{n+1}-1)$$

Now if I assume that $a,b \in \mathbb{N}$ and $a\neq b$ otherwise is a trivial solution. Also that $(a-1)\neq p_1$,$(a-1)\neq p_2$, clearly:

$$(a-1)|(b^{n}-1)$$ $$(a-1)|(b^{n+1}-1)$$

In other terms:

$$b^{n}\equiv 1 \mod (a-1) \tag{1}$$ $$b^{n+1}\equiv 1 \mod (a-1)\tag{2}$$

Using some properties of modular arithmetic in (1):

$$b^{n+1}\equiv b \mod (a-1) $$ $$b\equiv b^{n+1} \mod (a-1) $$

Now using (2) $b\equiv b^{n+1} \mod (a-1) $ and $b^{n+1}\equiv 1 \mod (a-1)$ :

$$b\equiv 1 \mod (a-1) $$

This means, $(a-1)|(b-1)$ and we also have $(b-1)|(a-1)$, now as $a,b \in \mathbb{N}$

$$a-1=b-1 \Rightarrow a=b$$

Contradiction !!

Our assumption that both are primes is wrong, so at least one must be prime.

** The other cases when $(a-1)= p_1$ or $(a-1)= p_2$ are not proven. Maybe you can work them out