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.
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.
$\displaystyle a\mid b^2\Longrightarrow b^2=xa\,,\,b^2\mid a^3\Longrightarrow a^3=yb^2\Longrightarrow$ 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
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 $\displaystyle p^\alpha\mid a\,,\,\,p^\beta\mid b$ and these powers are maximal in each case, i.e. $\displaystyle p^{\alpha+1}\nmid a\,,\,\,p^{\beta+1}\nmid b$, and suppose $\displaystyle \alpha\leq \beta$ , then:
$\displaystyle a\mid b^2\Longrightarrow b^2=xa$ . But $\displaystyle p^{2\beta}\mid b^2\Longrightarrow p^{2\beta}\mid xa\Longrightarrow p^{2\beta-\alpha}\mid x$ . Continue from here comparing odd powers of a with even powers of b.
Tonio