Is the following system of equations solvable, and also the one that arises from it by taking 8 instead of 4 as the base on the right side of the first equation? If there is a solution, let's find it; if we arrive at a negative answer, let's provide the explanation.

$$
\begin{aligned}
& 2^{x}+2^{x+1}+2^{x+2}+2^{x+3}=4^{y}+4^{y+1}+4^{y+2}+4^{y+3} \\
& 3^{x}+3^{x+1}+3^{x+2}+3^{x+3}=9^{y}+9^{y+1}+9^{y+2}+9^{y+3}
\end{aligned}
$$
By factoring and considering $4=2^{2}$ and $9=3^{2}$, the first equation can be written as:

$$
2^{x}\left(1+2+2^{2}+2^{3}\right)=2^{2 y}\left(1+4+4^{2}+4^{3}\right)
$$

from which, by division,

$$
2^{x-2 y}=85 / 15=17 / 3
$$

Similarly, from the second equation,

$$
40 \cdot 3^{x}=820 \cdot 3^{2 y}, \quad 3^{x-2 y}=41 / 2
$$

Taking the logarithm of both sides from (4) and (5),

$$
x-2 y=\frac{\lg 17-\lg 3}{\lg 2} \approx 2.5, \quad \text { or } \quad x-2 y=\frac{\lg 41-\lg 2}{\lg 3} \approx 2.7
$$

The left sides are equal, but the right sides are different; our system of equations is contradictory, hence it has no solution.

If on the right side of (1) the base 4 is replaced with $8=2^{3}$, similarly to the above transformations,

$$
15 \cdot 2^{x}=585 \cdot 2^{3 y}, \quad 2^{x-3 y}=39
$$

From the exponent, we see that the system of linear equations that can be obtained from here and (5) will be solvable for $x$ and $y$. Calculating to 4 significant figures,

$$
x-3 y \approx 5.285, \quad x-2 y \approx 2.749
$$

by subtraction and rounding to 3 significant figures,

$$
y \approx-2.54, \quad \text { and thus } \quad x \approx-2.32
$$

With these, the two sides of (5), or the modified (1) are $40 \cdot 3^{x} \approx 3.128$ and $820 \cdot 9^{y} \approx 3.090$, or $15 \cdot 2^{-2.32} \approx 3.005$ and $585 \cdot 8^{-2.54} \approx 2.973$.

Ernő Kovács (Pécs, Széchenyi I. g. III. o. t.)