[b]p1. [/b](a) Show that for every positive integer $m > 1$, there are positive integers $x$ and $y$ such that $x^2 - y^2 = m^3$. 
(b) Find all pairs of positive integers $(x, y)$ such that $x^6 = y^2 + 127$. 


[b]p2.[/b] (a) Let $P(x)$ be a polynomial with integer coefficients. Suppose that $P(0)$ is an odd integer and that $P(1)$ is also an odd integer. Show that if $c$ is an integer then $P(c)$ is not equal to $0$. 
(b) Let P(x) be a polynomial with integer coefficients. Suppose that $P(1,000) = 1,000$ and $P(2,000) = 2,000.$ Explain why $P(3,000)$ cannot be equal to $1,000$. 


[b]p3.[/b] Triangle $\vartriangle ABC$ is created from points $A(0, 0)$, $B(1, 0)$ and $C(1/2, 2)$. Let $q, r$, and $s$ be numbers such that $0 < q < 1/2 < s < 1$, and $q < r < s$. Let D be the point on $AC$ which has $x$-coordinate $q$, $E$ be the point on AB which has $x$-coordinate $r$, and $F$ be the point on $BC$ that has $x$-coordinate $s$.
(a) Find the area of triangle $\vartriangle DEF$ in terms of $q, r$, and $s$. 
(b) If $r = 1/2$, prove that at least one of the triangles $\vartriangle ADE$, $\vartriangle CDF$, or $\vartriangle BEF$ has an area of at least $1/4$.

[b]p4.[/b] In the Gregorian calendar:
(i) years not divisible by $4$ are common years,
(ii) years divisible by $4$ but not by $100$ are leap years,
(iii) years divisible by $100$ but not by $400$ are common years,
(iv) years divisible by $400$ are leap years,
(v) a leap year contains $366$ days, a common year $365$ days.
From the information above:
(a) Find the number of common years and leap years in $400$ consecutive Gregorian years. Show that $400$ consecutive Gregorian years consists of an integral number of weeks. 
(b) Prove that the probability that Christmas falls on a Wednesday is not equal to $1/7$.


[b]p5.[/b] Each of the first $13$ letters of the alphabet is written on the back of a card and the $13$ cards are placed in a row in the order $$A,B,C,D,E, F, G,H, I, J,K, L,M$$
The cards are then turned over so that the letters are face down. The cards are rearranged and again placed in a row, but of course they may be in a different order. They are rearranged and placed in a row a second time and both rearrangements were performed exactly the same way. When the cards are turned over the letters are in the order $$B,M, A,H, G,C, F,E,D, L, I,K, J$$ What was the order of the letters after the cards were rearranged the first time? 



PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
### Part (a)
We need to show that for every positive integer \( m > 1 \), there exist positive integers \( x \) and \( y \) such that \( x^2 - y^2 = m^3 \).

1. Start with the identity for the difference of squares:
   \[
   x^2 - y^2 = (x + y)(x - y)
   \]
2. Set \( x + y = m^2 \) and \( x - y = m \). This gives us a system of linear equations:
   \[
   \begin{cases}
   x + y = m^2 \\
   x - y = m
   \end{cases}
   \]
3. Solve this system for \( x \) and \( y \):
   \[
   x = \frac{(x + y) + (x - y)}{2} = \frac{m^2 + m}{2}
   \]
   \[
   y = \frac{(x + y) - (x - y)}{2} = \frac{m^2 - m}{2}
   \]
4. Verify that \( x \) and \( y \) are positive integers:
   - Since \( m > 1 \), both \( m^2 + m \) and \( m^2 - m \) are even, ensuring \( x \) and \( y \) are integers.
   - Both \( x \) and \( y \) are positive because \( m^2 > m \).

Thus, \( x = \frac{m^2 + m}{2} \) and \( y = \frac{m^2 - m}{2} \) are positive integers satisfying \( x^2 - y^2 = m^3 \).

### Part (b)
We need to find all pairs of positive integers \( (x, y) \) such that \( x^6 = y^2 + 127 \).

1. Let \( X = x^3 \). Then \( x^6 = X^2 \), and the equation becomes:
   \[
   X^2 = y^2 + 127
   \]
2. This is a Diophantine equation of the form \( X^2 - y^2 = 127 \), which can be factored as:
   \[
   (X + y)(X - y) = 127
   \]
3. Since 127 is a prime number, the factor pairs of 127 are \( (1, 127) \) and \( (-1, -127) \). We consider the positive pair:
   \[
   \begin{cases}
   X + y = 127 \\
   X - y = 1
   \end{cases}
   \]
4. Solve this system for \( X \) and \( y \):
   \[
   X = \frac{(X + y) + (X - y)}{2} = \frac{127 + 1}{2} = 64
   \]
   \[
   y = \frac{(X + y) - (X - y)}{2} = \frac{127 - 1}{2} = 63
   \]
5. Since \( X = x^3 \), we have:
   \[
   x^3 = 64 \implies x = 4
   \]

Thus, the pair \( (x, y) \) is \( (4, 63) \).

The final answer is \( \boxed{ (4, 63) } \)