If $a\mid b^2, b^2\mid a^3,\ldots ,a^n\mid b^{n+1},b^{n+1}\mid a^{n+2},\ldots$ then $a=b$

Question

I'm stuck with this problem :

Let $a,b$ positive integers such that $$a\mid b^2, b^2\mid a^3,\ldots ,a^n\mid b^{n+1},b^{n+1}\mid a^{n+2},\ldots$$ Show that $a=b$.

If were $ b > a $ then $\lim_{n \to \infty}\frac{b^n}{a^n}=0 $ choosing $\epsilon = \frac{1}{a}$ we got a contradiction, but I can't show that $b<a$ can't hold.

Any help is apreciated.

Answer 1

Assume by contradiction $a \neq b$. Then either $a<b$ or $a>b$.

Case 1: If $a <b$ then as $b^{2n}|a^{2n+1}$ we have $b^{2n} \leq a^{2n+1}$.

Therefore $$(\frac{b}{a})^{2n} \leq a$$

But since $\frac{b}{a} >1$ we also have $$\lim_n (\frac{b}{a})^{2n} = \infty$$ This is a contradiction.

Case 2: If $b <a$ then as $a^{2n-1}|b^{2n}$ we have $a^{2n-1} \leq b^{2n}$.

Therefore $$(\frac{a}{b})^{2n-1} \leq b$$

But since $\frac{a}{b} >1$ we also have $$\lim_n (\frac{a}{b})^{2n-1} = \infty$$ This is a contradiction.

Answer 2

Let $v_p(x)$ denote the power of a prime $p$ in the factorization of $x$ .

Let $p$ be a prime with $p \mid a$ so it follows that $p \mid b$ (from $a \mid b^2$ ) . The reverse also holds so $a$ and $b$ must have the same prime factors .

Now take such a prime $p$ .

For every $k \geq 1$ we know that :

$$a^{2k-1} \mid b^{2k}$$ and we can deduce that :

$$v_p(a^{2k-1}) \leq v_p(b^{2k})$$

It's not hard to see that in general $v_p(x^y)=y v_p(x)$ so after rearranging :

$$\frac{v_p(a)}{v_p(b)} \leq \frac{2k}{2k-1}$$

But this holds for every $k$ so taking $k \to \infty$ we get :

$$\frac{v_p(a)}{v_p(b)} \leq 1$$

Using the other relations (of the form $b^{2k} \mid a^{2k+1}$ ) we can deduce that :

$$\frac{v_p(a)}{v_p(b)} \geq 1$$

It follows then that $v_p(a)=v_p(b)$ and this must hold for every prime that divides them . Thus the conclusion $a=b$ must hold .

Answer 3

Pick a prime $p$, we must show $v_p(a)=v_p(b)$.

From the divisibility relations we have, for each $n\in \mathbb Z^+$:

$v_p(a)\leq \frac{v_p(b)2n}{2n-1}\implies v_p(a)\leq v_p(b)$.

We also have:

$v_p(b)\leq \frac{v_p(a)2n+1}{2n}\implies v_p(b)\leq v_p(a)$

Answer 4

Let $p$ be a prime. Assume $p^k\|a$ (i.e., $p^k\mid a$ and $p^{k+1}\nmid a$) and $p^l\|b$. Then from $a^n\mid b^{n+1}$ we get $nk\le (n+1)l$ and from $b^{n+1}\mid a^{n+2}$ we get $(n+1)l\le (n+2)k$. Hence $$ -k\le (n+1)(l-k)\le k$$ or $$ (n+1)|l-k|\le k.$$ For $n\gg k$ we obtain $|l-k|<1$ hene $l=k$. As this holds for all primes $p$ and $a,b$ are positive, we conclude $a=b$.