Exercise 13. For any real number $x$, we denote $\lfloor x\rfloor$ as the greatest integer less than or equal to $x$, and we call the fractional part of $x$ the number $\langle x\rangle$ defined as $\langle x\rangle=x-\lfloor x\rfloor$. How many real numbers $x$ satisfy $1 \leqslant x \leqslant 10$ and $\langle x\rangle^{2}=\left\langle x^{2}\right\rangle$?
Solution to Exercise 13: Let's try to express \( x^2 \) in terms of \( \lfloor x \rfloor + \langle x \rangle \) to calculate its fractional part. We write \( x = \lfloor x \rfloor + \langle x \rangle \), hence \( x^2 = (\lfloor x \rfloor)^2 + 2 \lfloor x \rfloor \cdot x + \langle x \rangle^2 \).

Thus, if \( \left\langle x^2 \right\rangle = \langle x \rangle^2 \), we get that \( \left\lfloor x^2 \right\rfloor = x^2 - \left\langle x^2 \right\rangle = (\lfloor x \rfloor)^2 + 2 \lfloor x \rfloor \cdot x \) is an integer, so \( 2 \lfloor x \rfloor \cdot x \) is also an integer. In particular, if we set \( k = \lfloor x \rfloor \), since \( k \geq 1 \), we know that \( x \) is of the form \( \frac{q}{2k} \). Since \( k \) is the smallest integer less than or equal to \( x \), we have \( k \leq x < k+1 \), thus \( 2k^2 \leq q < 2k^2 + 2k \). Let \( q = r + 2k^2 \) with \( 0 \leq r < 2k \), then \( x = k + \frac{r}{2k} \).

Conversely, suppose \( x \) is of the form \( k + \frac{r}{2k} \) with \( k \geq 1 \) and \( 0 \leq r < 2k \), we get that \( k \leq x < k+1 \), so \( \lfloor x \rfloor = k \), thus \( \langle x \rangle = \frac{r}{2k} \).

Furthermore, \( x^2 = k^2 + r + \frac{r^2}{(2k)^2} \) satisfies \( k^2 + r \leq x^2 < k^2 + r + 1 \), so \( \left\lfloor x^2 \right\rfloor = k^2 + r \) which gives \( \left\langle x^2 \right\rangle = \frac{r^2}{(2k)^2} = \langle x \rangle^2 \): \( x \) is a solution.

It remains to count how many such \( x \) are between 1 and 10: since \( k = \lfloor x \rfloor \), we must have \( 1 \leq k \leq 10 \). For \( 1 \leq k \leq 9 \), we have \( 1 \leq k \leq k + \frac{r}{2k} < k+1 \leq 10 \), so \( x \) is indeed between 1 and 10. Therefore, for a fixed \( k \), there are \( 2k \) solutions, so we have \( 2(1 + 2 + \cdots + 8 + 9) = 90 \) solutions with \( 1 \leq k \leq 9 \). If \( k = 10 \), \( x \geq 10 \) so \( x = 10 \) which is indeed a solution (this corresponds to \( k = 10, r = 0 \)). Therefore, there are a total of 91 solutions.