[b]p1.[/b] Every point of the plane is painted either red or green, and there is at least one red point and at least one green point. Prove that
a) There exist two points painted the same color as each other and lying exactly $1$ cm apart.
b)  There exist two points painted different colors from each other, and lying exactly $1$ cm apart.


[b]p2.[/b] Two people play the following game: there are two bunches of matches. The first bunch contains $m$ matches, the second bunch contains $n$ matches, $m$ and $n$ are not zero. On his turn each player takes as many matches from only one bunch as he wants, but he must take at least one match. The loser is the player who takes the last match. Prove that if $m$ is not equal to $n$, then the first player can win regardless of the strategy of the second player.


[b]p3.[/b] Suppose you are given some numbers whose sum is $1$. Is it possible for the sum of their squares to be less than  one tenth?

[b]p4.[/b] Suppose  $p$ is a prime number. Show that the number $1+\frac12+\frac13+\frac14+...+\frac{1}{p}$ is not an integer.


PS. You should use hide for answers.
### Problem 1a
1. Consider an equilateral triangle with side length 1 cm.
2. Each vertex of this triangle is a point in the plane.
3. Since there are only two colors (red and green) and three vertices, by the pigeonhole principle, at least two of these vertices must be the same color.
4. Therefore, there exist two points painted the same color that are exactly 1 cm apart.

\(\blacksquare\)

### Problem 1b
1. Consider a point \(A\) painted red and a point \(B\) painted green.
2. Draw a circle of radius 1 cm centered at point \(A\).
3. Since the plane is infinite and every point is either red or green, there must be at least one point on this circle.
4. If this point is red, then we have two red points 1 cm apart.
5. If this point is green, then we have a red point and a green point 1 cm apart.
6. Therefore, there exist two points painted different colors that are exactly 1 cm apart.

\(\blacksquare\)

### Problem 2
1. Let \(m\) and \(n\) be the number of matches in the two bunches, with \(m \neq n\).
2. The first player can always ensure that after their turn, the two bunches have an equal number of matches.
3. If \(m > n\), the first player takes \(m - n\) matches from the first bunch, making both bunches have \(n\) matches.
4. If \(n > m\), the first player takes \(n - m\) matches from the second bunch, making both bunches have \(m\) matches.
5. From this point on, the first player can always mirror the second player's moves, ensuring that the second player is forced to take the last match.
6. Therefore, the first player can always win if \(m \neq n\).

\(\blacksquare\)

### Problem 3
1. Let the numbers be \(x_1, x_2, \ldots, x_n\) such that \(\sum_{i=1}^n x_i = 1\).
2. We need to check if \(\sum_{i=1}^n x_i^2 < \frac{1}{10}\).
3. By the Cauchy-Schwarz inequality, \((\sum_{i=1}^n x_i^2)(\sum_{i=1}^n 1) \geq (\sum_{i=1}^n x_i)^2\).
4. Since \(\sum_{i=1}^n 1 = n\) and \(\sum_{i=1}^n x_i = 1\), we have \(n \sum_{i=1}^n x_i^2 \geq 1\).
5. Therefore, \(\sum_{i=1}^n x_i^2 \geq \frac{1}{n}\).
6. For \(\sum_{i=1}^n x_i^2 < \frac{1}{10}\), we need \(\frac{1}{n} < \frac{1}{10}\), which implies \(n > 10\).
7. Thus, it is possible for the sum of their squares to be less than one tenth if there are more than 10 numbers.

\(\boxed{\text{True}}\)

### Problem 4
1. Consider the harmonic sum \(H_p = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p}\).
2. Express \(H_p\) as a single fraction: \(H_p = \frac{a}{b}\), where \(b = p!\).
3. The numerator \(a\) is the sum of the terms \(\frac{p!}{1}, \frac{p!}{2}, \ldots, \frac{p!}{p}\).
4. The first \(p-1\) terms are divisible by \(p\), but the last term \(\frac{p!}{p} = (p-1)!\) is not divisible by \(p\).
5. Therefore, the numerator \(a\) is not divisible by \(p\), while the denominator \(b\) is divisible by \(p\).
6. Hence, \(H_p\) cannot be an integer.

\(\blacksquare\)