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Let $a,b$ be positive integers such that $a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots$ that is $a^{2n-1}\mid b^{2n} ; b^{2n}\mid a^{2n+1} , \forall n \in \mathbb Z^+$ , then is it true that $a=b$ ?

Martin Sleziak
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    If $a \neq b$ then there can be 2 options - either there is an uncommon prime divisor, or there is only a difference in prime powers. Think about what this means for the powers of prime divisors... – mousomer Feb 11 '15 at 14:15
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    This might be useful http://math.stackexchange.com/questions/704135/number-theory-why-is-my-approach-incorrect – kingW3 Feb 11 '15 at 14:16
  • @mousomer : off-course I can clearly see that $a,b$ has same set of prime divisors , so only difference is in powers of those primes –  Feb 11 '15 at 14:17
  • So it's really a question of simple arithmetic. If the powers are $x$ and $y$ then what this amounts to is $2ny-y \le 2nx \le 2ny+y$ – mousomer Feb 11 '15 at 14:20
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    See this – Bart Michels Feb 11 '15 at 14:21
  • See a recent duplicate with answers: http://math.stackexchange.com/questions/1281226/given-positive-integers-a-and-b-such-that-a-mid-b2-b2-mid-a3-a3-mid-b – user26486 May 26 '15 at 04:02

4 Answers4

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Let $p$ a prime that divides $a$ then since $a|b^2$ then $p$ divides $b$. Similarly we get that if $p$ is a prime that divides $b$ then it divides $a$ hence $a$ and $b$ has the same primes in their primary decomposition. Now if $p^\alpha$ is the factor of the primary decomposition of $a$ and $p^\beta$ is that of $b$ then

$$\alpha\le2\beta\le3\alpha\le 4\beta\le\cdots$$ then

$$\alpha\le\frac {2n}{2n-1}\beta,\quad \forall n$$ and $$\beta \le\frac {2n+1}{2n}\alpha,\quad \forall n$$ so by taking the limit $n\to\infty$ we get $\alpha=\beta$ and then $a=b$.

Michael Hardy
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If $a <b$ then $$\lim_n\frac{b^{2n}}{a^{2n+1}}= \infty$$

which means that from some point $b^{2n} >a^{2n+1}$, and it cannot divide it.

Same way, if $a>b$ then

$$\lim_n\frac{a^{2n-1}}{b^{2n}}= \infty$$

N. S.
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This is a twist of another answer here in order to avoid prime factorization.

We may assume $a,b>1$. Then $$\frac{2n-1}{2n} \leq \frac{\log b}{\log a} \leq \frac{2n+1}{2n}$$ for all $n$. Taking $n \to \infty$, we get $a=b$.

Bart Michels
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Under the given condition, it is clear that the prime factors of $a$ and $b$ are the same. Without loss of generality, you can assume that $a = p^i$ and $b=p^j$ where $p$ is a prime factor and $i,j\in \mathbb N$ (for if $a_k$ and $b_k$ are of this form and fulfill the given condition, it is clear that $a$ = product of $a_k$ and $b$ = product of $b_k$ fulfills this condition too, and conversely etc.). There holds $p^i|p^{2j}$, $p^{2j}|p^{3i}$ etc. Hence $i \leq 2j \leq 3i \leq 4j \cdots \leq n i \leq (n+1) j\leq (n+2) i$ etc. Taking $n$ large shows that $i=j$.

In conclusion, yes $a = b$.

MikeTeX
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