\section*{Problem 3 - 151033}

When printing a math problem, instead of \(\left(1+a^{2} x^{2}\right): x^{2}=b\) (with given numbers \(a, b\)), the equation \(\left(1+a^{2} x^{2}\right) \cdot x^{2}=b\) (with the same numbers \(a, b\)) was printed by mistake.

Despite this, the resulting equation had the same non-empty solution set as the originally intended equation.

Determine this solution set!
}

We assume that \(a, b \in \mathbb{R}\) and \(x \in \mathbb{R} \backslash\{0\}\). From the first, "actually correct" problem, we have on the one hand:

\[
1+a^{2} x^{2}=b x^{2}
\]

The second, "incorrectly printed" problem, on the other hand, reads: (2) \(\quad\left(1+a^{2} x^{2}\right) \cdot x^{2}=b\). Substituting (1) into (2) yields:

\[
b x^{2} \cdot x^{2}=b
\]

While \(b=0\) satisfies this equation, it would require, due to (1), that

\[
1+a^{2} x^{2}=0
\]

which is not solvable in \(\mathbb{R}\). Therefore, in (3), it must hold that: \(x^{4}=1\). Since \(x \in \mathbb{R}\), the solution set is \(\{1,-1\}\).