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QUESTION :

Let $a,b$ be positive integers such that $a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots$.....so on, then prove that $a=b$

This is what I tried:

Say $a≠b$ rather than $a>b≥1$

$$\exists k :a^{k-1}>b^k$$ Then: $$a^{k-1}>b^k$$ $$\left(\frac{a}{b}\right)^k>a$$ $$k>\log_{\left(\frac{a}{b}\right)}a$$ I am struck after this. I don't know what to do next (even I don't know what I have done is right or wrong).

What to do next?

I apologise for asking a same question which exist already on S.E. but I didn't understand that(due to limit).

miracle173
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Wolgwang
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    If you don't understand some prior answers then please first ask for elaboration there before posting duplicate questions. If you receive no (sufficient) reply there then ask a more specific question on the point in the proof that was not clear (not the whole problem). – Bill Dubuque Sep 10 '20 at 15:57
  • A conceptual way to view this number theoretically it to employ the unboundedness of the denominators of powers of proper fractions, i.e. if a common denominator exists for a sequence of unbounded powers of a fraction then the fraction is integer. So here $,b(b/a),,b(b/a)^3,b(b/a)^5\ldots,$ are all integers, so $,b,$ is a common denom for all odd powers of $,b/a,$ so $,b/a\in\Bbb Z,$ so $,a\mid b.,$ Similarly $,b\mid a,$ so $,a = \pm b.\ \ \ $ – Bill Dubuque Sep 10 '20 at 16:55
  • If you chase all the "Linked" questions in the linked dupes you will find many answers at various levels of detail, including generalizations – Bill Dubuque Sep 10 '20 at 16:55
  • @Bill Dubuque I wanted to know, what I have done is right or wrong. – Wolgwang Sep 11 '20 at 00:55

2 Answers2

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What you have done is fine. You have shown that for $k > \log_{\frac ab}a$ (which is some constant), we will have:

$$a^{k-1}>b^k$$

For this (possibly very large) $k$, we cannot write down $$a^{k-1} \mid b^k$$

since a divisor of a positive integer cannot exceed that integer.

You have shown what they have done with limits by giving a concrete bound.

player3236
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This follows in a finite number of the division statements, if you use the prime factorization of both. If a prime divides one of them (the exponent of the prime is nonzero) then it divides the other, just from the first two statements

$$a\mid b^2 , b^2\mid a^3 .$$ Now let $m,n \geq 1$ be integers, the exponents in the prime factorizations, $$ a = p^m \cdot \mbox{OTHER_A} $$ $$ b = p^n \cdot \mbox{OTHER_B} $$

Let $a,b$ be positive integers such that $a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots$.....so on, then prove that $a=b$

For this prime $p$, the first few "divides" statements read

$$ m \leq 2n \; , \; $$ $$ 2n \leq 3m \; , \; $$

$$ 3m \leq 4n \; , \; $$ $$ 4n \leq 5m \; , \; $$

For any positive odd number $c$ we have the pair $$ cm \leq (1+c) n \; , \; $$

Take $c > n$ so that $$ \frac{n}{c} < 1. $$ $$ m \leq \left(1 + \frac{1}{c} \right) n = n + \frac{n}{c} < n+1. $$ Put together, $m < n+1$ and $$ m \leq n \; . $$

For even positive integer $d$ $$ d n \leq (1+d) m \; . \; $$ Take $d > m$ so that $$ \frac{m}{d} < 1. $$ $$ n \leq \left(1 + \frac{1}{d} \right) m = m + \frac{m}{d} < m+1. $$ Put together, $n < m+1$ and $$ n \leq m \; . $$

As $n \leq m$ and $m \leq n,$ we have $$ m=n $$

These were exponents of some prime $p$ that divides your two numbers. The argument applies to every prime that divides at least one of the numbers: the exponents are equal. Therefore $a=b.$

Will Jagy
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