Solve the following equation:

$$
1+a+a^{2}+\ldots+a^{2 x-1}+a^{2 x}=(1+a)\left(1+a^{2}\right)\left(1+a^{4}\right)\left(1+a^{8}\right)
$$
I. Solution: From the problem statement, it is clear that the exponents on the left side can only be positive integers, so \( x \) can only be positive. Let's break down the sequence on the right side:

$$
1+a+a^{2}+\ldots+a^{2 x-1}+a^{2 x}=1+a+a^{2}+\ldots+a^{14}+a^{15}
$$

Two ordered polynomials can only be identically equal if their terms match term by term, so if

$$
a^{2 x}=a^{15}
$$

(if this holds, then the other terms will also match). If \( a \neq \pm 1 \) or 0, it follows from the obtained equality that

$$
2 x=15, \quad x=7.5
$$

If \( a=1 \), the value on the right side is 16, so the equality can only hold if \( 2 x=15, x=7.5 \), as before. If \( a=-1 \), the right side is 0, which can only happen if \( 2 x \) is any odd number, i.e., \( x=\frac{2 k+1}{2}(k=0,1,2, \ldots) \). If finally \( a=0 \), then 1 stands on both sides, so \( 2 x \) can be any positive integer, i.e., \( x=\frac{l}{2}(l=0,1,2,3 \ldots) \).

Mocskónyi Miklós (Bp. V, Eötvös g. II. o. t.)

II. Solution: On the left side, there is a geometric sequence with ratio \( a \) and \( 2 x+1 \) terms. If \( a \neq 1 \), we can use the formula for the sum of a geometric sequence:

$$
\frac{a^{2 x+1}-1}{a-1}=(1+a)\left(1+a^{2}\right)\left(1+a^{4}\right)\left(1+a^{8}\right)
$$

Multiply through by \( (a-1) \) and repeatedly use the identity for the product of the sum and difference of two numbers:

$$
\begin{gathered}
a^{2 x+1}-1=\left(a^{2}-1\right)\left(1+a^{2}\right)\left(1+a^{4}\right)\left(1+a^{8}\right)=\left(a^{4}-1\right)\left(a^{4}+1\right)\left(a^{8}+1\right)= \\
=\left(a^{8}-1\right)\left(1+a^{8}\right)=a^{16}-1
\end{gathered}
$$

From this

$$
a^{2 x+1}=a^{16}
$$

For \( a \neq \pm 1,0 \), it follows that

$$
x=7.5
$$

The analysis of the excluded cases can be carried out in the same way as in the first solution.

Mezey Ferenc (Bp. II, Rákóczi g. II. o. t.)