1. Determine all positive integers $x$, such that $2x+1$ is a perfect square, but among the numbers $2x+2, 2x+3, \cdots, 3x+2$, there is no perfect square.
Solution. Let $n$ be an integer such that $2 x+1=n^{2}$ and $n^{2} \leq 3 x+2<(n+1)^{2}$. From the first equation, we get $x=\left(n^{2}-1\right) / 2$, and substituting this value into the double inequality, we obtain

$$
n^{2} \leq \frac{3 n^{2}+1}{2}<n^{2}+2 n+1 \Leftrightarrow 2 n^{2} \leq 3 n^{2}+1<2 n^{2}+4 n+2
$$

In the last expression, the first inequality holds for all integer $n$, the second can be written as

$$
n^{2}-4 n-1<0 \Leftrightarrow 2-\sqrt{5}<n<2+\sqrt{5}
$$

and is satisfied for all integers $n \in\{0,1,2,3,4\}$. Among these values, only for $n=3$, we get that $x=\left(3^{2}-1\right) / 2=4$ is a positive integer, and this is the only solution to the problem, since in this case $2 x+1=9$ and among the numbers $10,11,12,13$ and 14 there is no perfect square.

2. $2 n+1$ chips, white and black, are placed in a row $(n \geq 1)$. A chip is said to be balanced if the number of white chips to its left, plus the number of black chips to its right is $n$. Determine, with reasoning, whether the number of chips that are balanced is even or odd.