Which of the fractions have a value of $\frac{45}{56}$ and whose numerators are square numbers, and denominators are cube numbers? Among these fractions, which one has the simplest form?
45 and 56 are relatively prime numbers, and all fractions whose value is $\frac{45}{56}$,

$$
\frac{45 m}{56 m} \quad \text { or } \quad \frac{3^{2} \cdot 5 \cdot m}{2^{3} \cdot 7 \cdot m}
$$

are of this form. The numerator can only be a perfect square if all prime factors have even exponents; thus, it is necessary that $m=5 r^{2}$.

The denominator can only be a perfect cube if all prime factors have exponents that are multiples of 3; thus, it is necessary that $m=7^{2} \cdot s^{3}$. Consequently,

$$
m=5 r^{2}=7^{2} s^{3}
$$

it follows that $r$ is divisible by 7, and $s$ is divisible by 5, and thus $m$ is a multiple of $7^{2}$ and $5^{3}$:

$$
m=7^{2} \cdot 5^{3} \cdot t
$$

The fractions in question are:

$$
\frac{45 m}{56 m}=\frac{3^{2} \cdot 5^{4} \cdot 7^{2} \cdot t}{2^{3} \cdot 7^{3} \cdot 5^{3} \cdot t}=\frac{\left(3 \cdot 5^{2} \cdot 7\right)^{2} t}{(2 \cdot 7 \cdot 5)^{3} t}
$$

where $t$ must be both a perfect square and a perfect cube; this is only possible if $t=x^{6}$. Therefore, the sought fractions are:

$$
\frac{\left(3 \cdot 5^{2} \cdot 7\right)^{2} x^{6}}{(2 \cdot 5 \cdot 7)^{3} x^{6}} \text { in form. }
$$

Among these, the simplest one appears when $x=1$.

For $x=1$,

$$
\frac{45}{56}=\frac{45 \cdot 5^{3} \cdot 7^{2}}{56 \cdot 5^{3} \cdot 7^{2}}=\frac{45 \cdot 6125}{56 \cdot 6125}=\frac{(525)^{2}}{70^{3}}
$$

Pál Koren (Fazekas Mihály g. VI. o., Debrecen.)