\section*{Problem 3A - 151043A}

If \(z\) is a real number, let \([z]\) denote the integer \([z]=g\) such that \(g \leq z < g+1\).

Determine all real numbers \(x\) for which \(-10 \leq x \leq 2\) and \(\left[x^{2}\right]=[x]^{2}\) hold!
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Obviously, \([x]\) rounds the real number \(x\) down to the nearest integer. According to the definition, the following holds:

\[
\left[x^{2}\right] \leq x^{2} < \left[x^{2}\right] + 1
\]

Given \(f = [x]\) and \(g = [x^2]\), the inequality can be rewritten as:

\[
g \leq x^2 < g + 1
\]

Since \(f \leq x < f + 1\), it follows that:

\[
f^2 \leq x^2 < (f + 1)^2
\]

Combining these, we get:

\[
f^2 \leq x^2 < g + 1 \leq (f + 1)^2
\]

This implies:

\[
f^2 \leq g < (f + 1)^2
\]

Since \(g\) is an integer, we have:

\[
f^2 \leq g \leq f^2 + 2f
\]

Thus, \(g\) can take the values \(f^2, f^2 + 1, \ldots, f^2 + 2f\).

For \(f = 0\), \(g\) can be \(0\).

For \(f = 1\), \(g\) can be \(1\) or \(2\).

For \(f = -1\), \(g\) can be \(1\).

For \(f = -2\), \(g\) can be \(4\).

For \(f = -3\), \(g\) can be \(9\).

For \(f = -4\), \(g\) can be \(16\).

For \(f = -5\), \(g\) can be \(25\).

For \(f = -6\), \(g\) can be \(36\).

For \(f = -7\), \(g\) can be \(49\).

For \(f = -8\), \(g\) can be \(64\).

For \(f = -9\), \(g\) can be \(81\).

For \(f = -10\), \(g\) can be \(100\).

Since \(2 g f + f^2 > 0\) and \(2 g f + f^2 < 1\) must hold, we can split this into two inequalities:

\[
\text { (1) } \quad 2 g f + f^2 > 0 \quad \text { and } \quad \text { (2) } \quad 2 g f + f^2 < 1
\]

From (1), since \(f^2 \geq 0\), it follows that:

\[
2 g f > -f^2
\]

Since \(f \neq 0\), we can divide by \(f\):

\[
2 g > -f
\]

This implies:

\[
2 g + f > 0
\]

From this, we can conclude that \(g \geq 0\) must hold, as \(f \neq 0\). Therefore, \(g\) can only be \(0\) or \(1\). From (2):

\[
-g - \sqrt{g^2 + 1} < f < -g + \sqrt{g^2 + 1}
\]

Since \(2 g f + f^2 < 1\) must hold, and the left side here is less than zero, a sharper condition is:

\[
0 < f < -g + \sqrt{g^2 + 1} \quad \text { or } \quad g < x < \sqrt{g^2 + 1}
\]

For \(g = 0\), this means \(0 < x < 1\), and for \(g = 1\), it follows that \(1 < x < \sqrt{2}\).

General solution: Combining these solution intervals with the integer solutions \(x = 0\) and \(x = 1\), we get \(0 \leq x < \sqrt{2}\), as well as all integers in the interval \([-10, 2]\).