Proving $\,a\mid b^2\mid a^3\mid b^4\mid a^5\ldots\implies a=b$.

Question

https://math.stackexchange.com/questions/704048/theory-number-problems After I saw that post i wanted to solve the first one which is $a\mid b^2,b^2\mid a^3,a^3\mid b^4,b^4\mid a^5\cdots$ Prove that $a=b$

Now i started by proving that $a$ and $b$ have same prime divisors,proving that is trivial,after doing so I checked do those factors have to have the same power.I tried with $$a=p^2,b=p^3$$ I noticed it exactly goes for 3 terms,or that $a^3\mid b^4$ doing $$a=p^{n-1},b=p^n\\p^{n-1}\mid p^{2n}\\p^{2n}\mid p^{3n-3}\\p^{3n-3}\mid p^{4n}\\p^{4n}\mid p^{5n-5}\\\cdots\\p^{n(n-1)}\mid p^{n(n-1)}\\p^{n(n-1)}\mid p^{n(n+1)}\\p^{n(n+1)}\not\mid p^{(n-1)(n+1)}$$ Now yeah I guess that would be a proof,but wouldn't setting $n=n+1$ infinitely many times make every term dividable?

For every prime that divides $;a;$, look at the maximal power of it that still divides it and show the same power divides $;b;$...and the other way around, of course. — DonAntonio, Mar 08 '14 at 13:48

Bill Dubuque · Accepted Answer · 2020-11-16T00:43:51.540

3

Hint $\ b(b/a),\,b(b/a)^3,b(b/a)^5\ldots\,$ are all integers, i.e. $\,b\,$ is a common denominator for $\rm\color{#c00}{unbounded}$ powers of $\,b/a,\,$ so $\,b/a,$ is an integer (prove it!), thus $\,a\mid b.\,$ Similarly $\,b\mid a.\,$

Hint $2\!:$ $\ $ Let $\, r = \frac{b}a = \frac{c}d,\ \color{#0a0}{(c,d)=1}.\,$ Then $\,b\:\!r^k = n_k\in\Bbb Z\,\Rightarrow\, b\:\!c^k = n_k d^k\,$ hence we infer by $\rm\color{#0a0}{Euclid}$: $\,d^k\mid b c^k\Rightarrow\, d^k\mid b^{\phantom{|^{|^|}}}\!\!$ for $\,\rm\color{#c00}{unbounded}$ $\,k\,$ so $\,d=\pm1,\,$ so $\,r = \frac{c}d = \pm c\in\Bbb Z,\,$ so $\,a\mid b$.

Remark $ $ That unbounded powers of proper fractions have no common denominator (such as $b$ above), is a special property of $\,\Bbb Z\,$ that needn't be true in general domains. When it holds true, a domain is called completely integrally closed.

See this answer for a generalization to Noetherian integrally closed domains, e.g. PIDs.

edited Nov 16 '20 at 00:43

answered Aug 10 '14 at 16:11

Bill Dubuque

272,048

“$b$ is a common denominator for unbounded powers of $b/a$, so $b/a$, is an integer.” What does unbounded power mean? – Wolgwang Sep 11 '20 at 01:06
@Mayank Unbounded powers of $r$ means $,r^{k_1},,r^{k_2},\ldots $ for an unbounded sequence of integers $,k_i.\ $ In the OP it is the sequence of all odd powers $,r, r^3, r^5,,\ldots\ \ $ – Bill Dubuque Sep 11 '20 at 01:18
Can you recommend some books for this particular concept? I haven't studied it till now. – Wolgwang Sep 11 '20 at 01:20
@Mayank Which concept? – Bill Dubuque Sep 11 '20 at 01:20
Unbounded powers. I need some more information about it. – Wolgwang Sep 11 '20 at 01:22
1

@Mayank It's just a set of powers of $r$ whose exponents are not bounded, e.g. all odd powes of $r$. The point is that denominators $d$ in this sequence will also be unbounded, so they can't all divide the integer $b$ as in the proof. – Bill Dubuque Sep 11 '20 at 01:27
I don't get what “not bounded”,“unbounded”,“bounded” mean? – Wolgwang Sep 11 '20 at 01:30
@Mayank A sequence of integers $,k_i,$ is unbounded if it has no upper bound, i.e. there is no integer $,m,$ such that all $,k_i < m\ \ $ – Bill Dubuque Sep 11 '20 at 01:32
Ok, I got it . Thanks a lot @Bill Dubuque – Wolgwang Sep 11 '20 at 01:34
A very classic approach – Ash_Blanc Jul 27 '23 at 11:34

score 2 · Answer 2 · answered Nov 16 '20 at 01:02

Let $\nu_p(n)$ be the $p$-adic valuation of $n$ for a prime $p$.

Note for each $n \in \mathbb{N}$, we have $(2n+1) \nu_p(a) \leq (2n+2) \nu_p(b)$ and $(2n+2) \nu_p(b) \leq (2n+3) \nu_p(a)$ so that $$1 - \frac{1}{2n+3} = \frac{2n+2}{2n+3} \leq \frac{\nu_p(a)}{\nu_p(b)} \leq \frac{2n+2}{2n+1} = 1+ \frac{1}{2n+1}.$$ As $n \to \infty$, $\frac{1}{2n+3} \to 0$ and $\frac{1}{2n+1} \to 0$, so $\frac{\nu_p(a)}{\nu_p(b)} \to 1$ and $\nu_p(a) = \nu_p(b)$ for each $p$. Thus, $a = b$ as desired.

Proving $\,a\mid b^2\mid a^3\mid b^4\mid a^5\ldots\implies a=b$.

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