Two square numbers add up to 148392. Which numbers are these?
If the sought numbers are $x$ and $y$, then

$$
x^{2}+y^{2}=148392=2^{3} \cdot 9^{2} \cdot 229
$$

where 229 is a prime number. $x$ and $y$ have the same parity. However, they cannot both be odd, because the sum of the squares of two odd numbers is only divisible by $2^{1}$, not by $2^{2}$. Therefore, both must be even, but they cannot both be multiples of $2^{2}$, because the sum of their squares is not divisible by $2^{4}$. Furthermore, it cannot be that one is a multiple of $2^{2}$ and the other a multiple of $2^{1}$, because if that were the case, the sum of their squares would be a multiple of $2^{2}$ and not of $2^{3}$.

If $x$ and $y$ are relatively prime to 3, then the sum of their squares would be a number of the form $3m+2$. Therefore, both $x$ and $y$ must have 3 as a prime factor. Since

$$
\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{3}\right)^{2}=9m
$$

$\frac{x}{3}$ and $\frac{y}{3}$ must both be multiples of 3, so $x$ and $y$ must be multiples of 9. Thus,

$$
x=2 \cdot 9a \quad \text{and} \quad y=2 \cdot 9b, \quad a^{2}+b^{2}=2 \cdot 229=458
$$

where $a$ and $b$ are odd and relatively prime numbers. In the squares of odd numbers, the units digit can only be $1, 5, 9$; since the units digit in the sum of two such squares is 8, the units digit in both squares must be 9, so the units digits of $a$ and $b$ must be 3 or 7.

However, $a^{2} < 458$. Indeed, $17^{2}+13^{2}=458$.

The sought numbers are: $2 \cdot 9 \cdot 13=234$ and $2 \cdot 9 \cdot 17=306$.

György Sommer and Sándor Grünfeld (Áll. Dobó István r. VI. o. Eger).