Prove x = y if $x | y^2$ and $y^2 | x^3$ and $x^3 | y^4$ ...

Question

So, I'm having trouble doing this certain problem. As the title suggests I should prove that if this holds:

$$x | y^2 $$ $$y^2 | x^3$$ $$x^3 | y^4$$ $$...$$

Than for each natural x and y, x is equal to y. I've tried writing the numbers x and y in form of:

$$x = p_1^{\alpha_1} + p_2^{\alpha_2} + ... + a_n^{\alpha_n}$$ $$y = P_1^{\beta_1} + p_2^{\beta_2} + ... + p_n^{\beta_n}$$

Now when I try to express the given equations in this form I get:

$$x \le y^2 \le x^3 \le y^4$$ Which for each $i$ yields:

$$\alpha_i \le 2\beta_i \le 3\alpha_i \le 4\beta_i ...$$

I was now wondering what to do next to prove it. Thanks.

Answer 1

It suffice to prove $\nu_p(x) = \nu_p(y)$ for all prime $p$.

We have the following $\nu_p$ information for any prime $p$ and positive integer $k$,

$(2k+1)\nu_p(x) \le (2k+2) \nu_p(y)$ and $(2k+2) \nu_p(y) \le (2k+3)\nu_p(x)$.

Which implies $\frac{2k+1}{2k+2}\nu_p(x)\leq \nu_p(y)\leq \frac{2k+3}{2k+2}\nu_p(x)$.

This implies $|\nu_p(x) - \nu_p(y)|$ is $0$, and so $\nu_p(x) = \nu_p(y)$ as desired.