$a|b^2$ , $b^2|a^3$ , $a^3|b^4$ , $b^4| a^5$... then...

Asked Oct 30 '18 at 01:04

Active Oct 30 '18 at 19:08

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Let $a$ and $b$ be positive interger such that $a|b^2$ , $b^2|a^3$ , $a^3|b^4$ , $b^4| a^5$ ... Prove that $a=b$.

Here how I tried to do it, I think it isn't the best way to approach it but anyway I failed in proving it:

EDIT: Let $a = p_1 ^{a_1} p_2 ^{a_2} ... p_k ^{a_k}$ and $b= p_1 ^{b_1} p_2 ^{b_2} ... p_k ^{b_k}$ and $a_i, b_i \geq 0$ for $i$, $ 0 \leq i \leq k$. Then for a specific prime $p_i$, $a_i \leq 2b_i$, $2b_i \leq 3a_i$ , $3a_i \leq 4b_i$ and $4b_i \leq 5a_i$.... I tried to get some information from that but didn't get anything.

This question has been marked as a duplicate but it is in fact different, or maybe that's what I can directly deduce from the conditions. Just look carefully at the titles to realise that they are different.

edited Oct 30 '18 at 19:08

asked Oct 30 '18 at 01:04

Vmimi

You don't get $a_i\mid 2b_i$ etc, but $a_i \le 2b_i$ etc. – hmakholm left over Monica Oct 30 '18 at 01:07
@Vmimi Though the exponent sequence differs slightly, the same methods in the linked dupe work here. – Bill Dubuque Oct 30 '18 at 01:23
@HenningMakholm You are right, I think I was just focused on division – Vmimi Oct 30 '18 at 01:42
@BillDubuque Sorry I edited it before seeing your comment. Thanks. – Vmimi Oct 30 '18 at 02:06
@BillDubuque Wouldn't this question be a better duplicate? – Arnaud D. Oct 30 '18 at 13:22
@ArnaudD. Yes, but I can't reopen and reclose. But it matters little since - as I said above - the method of solution is exactly the same for both. – Bill Dubuque Oct 30 '18 at 13:27
@BillDubuque Oh I thought the gold badge allowed you to do that, or maybe modify the "duplicate" notice somehow. But as you say, it doesn't really matter. – Arnaud D. Oct 30 '18 at 13:29
@ArnaudD. You can't close twice. In any case, the question is surely a dupe (and will likely be soon deleted) so there's little point to polish the dupe target. – Bill Dubuque Oct 30 '18 at 13:32
The title and the question doesn't match. Does the sequence stop at $b^4\mid a^5$ or does it go on forever? – user10354138 Oct 30 '18 at 18:25

$a|b^2$ , $b^2|a^3$ , $a^3|b^4$ , $b^4| a^5$... then...

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