If a and b are positive integers satisfying $(a^{([2n-1]^2)}) \vert b^{([2n]^2)}$ and $b^{([2n]^2)} \vert (a^{([2n+1]^2)})$ prove a = b.

Question

Recently encountered this question:

If a and b are positive integers satisfying $(a^{([2n-1]^2)})\ \vert \ b^{([2n]^2)}$ and $b^{([2n]^2)} \ \vert \ (a^{([2n+1]^2)})$ prove $a = b$.

My calculations so far:

Since $(a^{([2n-1]^2)})\ \vert \ b^{([2n]^2)}$ and $b^{([2n]^2)} \ \vert \ (a^{([2n+1]^2)})$

we have that $a^{(4n^2-4n+1)}\ \vert \ a^{(4n^2+4n+1)}$, thus there exists an integer $p$ such that

$p = a^{(4n^2+4n+1)}/a^{(4n^2-4n+1)}=a^{(4n^2+4n+1-4n^2+4n-1)}=a^{(8n)}$

$4n^2\log_{(b)}/\log_{(a)}-4n^2+4n-1+4n^2+4n+1-4n^2 \log_{(b)}/\log_{(a)}=8n$

I think they are wrong.

What should I do next? How would you attack this question (from the start, would you use my calculations)? Do you have an answer to this question? If possible, can you please explain it and show me the best way to do this?

Thanks a lot.

Edit: There has been some confusion with the question, here a photo:

Answer 1

Assume $a>b$, then $\log_ba>1$. Choose $n\in\mathbb N$ such that $$\log_ba>\left(\frac{2n}{2n-1}\right)^2$$ which is possible since the right side tends to $1$ for $n\to\infty$. Then $$\log_ba(2n-1)^2>(2n)^2\Longrightarrow a^{(2n-1)^2}>b^{(2n)^2}$$ and thus $a^{(2n-1)^2}\nmid b^{(2n)^2}$, which is a contradiction. You can deal with the case $b>a$ in a similar way.