# Thread: given positive integers a and b such that a|b^2, b^2|a^3, a^3|b^4,...

1. ## given positive integers a and b such that a|b^2, b^2|a^3, a^3|b^4,...

The problem below seems fairly easy... but I just can't manage to prove it!

PROBLEM: Given posistive integers a and b such that a|b^2, b^2|a^3, a^3|b^4, b^4|a^5, ... , prove that a = b.

2. Originally Posted by swallenberg
The problem below seems fairly easy... but I just can't manage to prove it!

PROBLEM: Given posistive integers a and b such that a|b^2, b^2|a^3, a^3|b^4, b^4|a^5, ... , prove that a = b.

$a\mid b^2\Longrightarrow b^2=xa\,,\,b^2\mid a^3\Longrightarrow a^3=yb^2\Longrightarrow$ this already tells us that every prime that divides a divides b and every prime that divides b divides a, so a,b are divisible by exactly the same primes. Well, now just prove that if a prime divides a to some power, then at that same power that primes divides b, and the other way around.

Tonio

3. Originally Posted by tonio
Well, now just prove that if a prime divides a to some power, then at that same power that primes divides b, and the other way around.

Tonio

But how? I came this far when I tried the first time... but I just can't seem to cross the finish line.

4. quote=swallenberg;446066]But how? I came this far when I tried the first time... but I just can't seem to cross the finish line.[/quote]

Suppose $p^\alpha\mid a\,,\,\,p^\beta\mid b$ and these powers are maximal in each case, i.e. $p^{\alpha+1}\nmid a\,,\,\,p^{\beta+1}\nmid b$, and suppose $\alpha\leq \beta$ , then:

$a\mid b^2\Longrightarrow b^2=xa$ . But $p^{2\beta}\mid b^2\Longrightarrow p^{2\beta}\mid xa\Longrightarrow p^{2\beta-\alpha}\mid x$ . Continue from here comparing odd powers of a with even powers of b.

Tonio

5. I finally realized how to do it.

(2n-1)alfa
≤ 2nbeta ≤ (2n+1)alfa, which is the same as (2n-1)/2n ≤ beta/alfa ≤ (2n+1)/2n.

Now if we let n tend to inf. we'll see that 1
beta/alfa ≤ 1 which means that alfa = beta.