Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$.

Question

Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$.

My (attempted) proof:

Suppose that $a^2 \mid b$ and $b^3 \mid c$, where $a$, $b$, and $c$ are integers.

Therefore, we have that $a^2 k_1 = b$ and $b^3 k_2 = c$ for some $k_1, k_2 \in \mathbb{Z}$.

I'm unsure of how to proceed from here. If I cube $a^2 k_1 = b$ to get $a^6 k_1 = b^3$, and then substitute this into $b^3 \mid c$, then I get $a^6k_1k_2 = c$, which isn't useful. The same happens if I do the other substitution.

I would greatly appreciate it if people could please take the time to clarify how I should be proceeding.

Answer 1

Using your notations: $a^2 k_1 = b$ and $b^3 k_2 = c$ for some $k_1, k_2 \in \mathbb{Z}$. Then $$c^3=b^9k_2^3=b^4b^5k_2^3=a^8k_1^4b^5k_2^3$$

Answer 2

I keep your notations and write $a^2 k_1 = b$ and $b^3 k_2 = c$. Then we have $$c^3=b^9k_2^3=b^2b^7k_2^3=(a^4k_1^2)b^7k_2^3$$ by squaring the first equation and replacing $b^2$. This can be written as $$c^3=(a^4b^5)k_3$$ where $k_3=b^2k_1^2k_2^3$ is an integer. Hence, $a^4b^5 \mid c^3$.

Answer 3

For some integers $k,l$, we have $a^2k =b, b^3l = c$.

Hence $c^3 = b^9l^3 = a^{18}k^9l^3 = (a^{10}k^5)a^8k^4l^3 = b^5a^4(a^4k^4l^3)$