## Task A-2.3.

Determine all quadruples $(a, b, c, d)$ of natural numbers such that

$$
a^{3}=b^{2}, \quad c^{5}=d^{4} \quad \mathrm{and} \quad a-c=9
$$
## Solution.

Numbers $a$ and $b$ must have the same prime factors. Let $p$ be one of their prime factors that appears in $a$ with exponent $k$, and in $b$ with exponent $l$. Then $p^{3 k}=p^{2 l}$, or $3 k=2 l$, from which it follows that $k$ is an even number.

Since this holds for all prime factors $p$, we conclude that $a$ is a perfect square, i.e., $a=n^{2}$ for some $n \in \mathbb{N}$.

1 point

Similarly, we conclude that $c=m^{4}$ for some $m \in \mathbb{N}$.

2 points

It follows that

$$
9=a-c=n^{2}-m^{4}=\left(n-m^{2}\right)\left(n+m^{2}\right)
$$

1 point

Since $n+m^{2}>0$, it follows that $n-m^{2}>0$.

Since $m$ is a natural number, $n-m^{2}>0$, the only possibility is $n+m^{2}=9, n-m^{2}=1$

from which it follows that $m=2$ and $n=5$.

It follows that $a=5^{2}=25, b=5^{3}=125, c=2^{4}=16, d=2^{5}=32$.