[b]p1.[/b] For what integers $n$ is $2^6 + 2^9 + 2^n$ the square of an integer?


[b]p2.[/b] Two integers are  chosen at random (independently, with repetition allowed) from the set $\{1,2,3,...,N\}$. Show that the probability that the sum of the two integers is even is not less than the probability that the sum is odd.


[b]p3.[/b] Let $X$ be a point in the second quadrant of the plane and let $Y$ be a point in the first quadrant. Locate the point $M$ on the $x$-axis such that the angle $XM$ makes with the negative end of the $x$-axis is twice the angle $YM$ makes with the positive end of the $x$-axis.


[b]p4.[/b] Let $a,b$ be positive integers such that $a \ge b \sqrt3$. Let $\alpha^n = (a + b\sqrt3)^n = a_n + b_n\sqrt3$ for $n = 1,2,3,...$.
i. Prove that $\lim_{n \to + \infty} \frac{a_n}{b_n}$ exists.
ii. Evaluate this limit.


[b]p5.[/b] Suppose $m$ and $n$ are the hypotenuses are of Pythagorean triangles, i.e,, there are positive integers $a,b,c,d$, so that $m^2 = a^2 + b^2$ and $n^2= c^2 + d^2$. Show than $mn$ is the hypotenuse of at least two distinct Pythagorean triangles.

Hint: you may not assume that the pair $(a,b)$ is  different from the pair $(c,d)$.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. We start with the equation given in the problem:
   \[
   2^6 + 2^9 + 2^n = k^2
   \]
   Simplifying the left-hand side, we get:
   \[
   2^6 + 2^9 + 2^n = 2^6 (1 + 2^3 + 2^{n-6}) = 2^6 (1 + 8 + 2^{n-6}) = 2^6 (9 + 2^{n-6})
   \]
   For this to be a perfect square, \(2^6 (9 + 2^{n-6})\) must be a perfect square. Let:
   \[
   2^6 (9 + 2^{n-6}) = k^2
   \]
   Dividing both sides by \(2^6\), we get:
   \[
   9 + 2^{n-6} = \left(\frac{k}{2^3}\right)^2
   \]
   Let \(m = \frac{k}{2^3}\), then:
   \[
   9 + 2^{n-6} = m^2
   \]
   Rearranging, we get:
   \[
   2^{n-6} = m^2 - 9
   \]
   Factoring the right-hand side, we get:
   \[
   2^{n-6} = (m-3)(m+3)
   \]
   Since \(2^{n-6}\) is a power of 2, both \(m-3\) and \(m+3\) must be powers of 2. Let:
   \[
   m-3 = 2^a \quad \text{and} \quad m+3 = 2^b
   \]
   where \(a < b\). Then:
   \[
   2^b - 2^a = 6
   \]
   Factoring out \(2^a\), we get:
   \[
   2^a (2^{b-a} - 1) = 6
   \]
   The only powers of 2 that satisfy this equation are \(2^1 \cdot (2^2 - 1) = 2 \cdot 3 = 6\). Thus, \(a = 1\) and \(b = 3\). Therefore:
   \[
   m-3 = 2^1 = 2 \quad \text{and} \quad m+3 = 2^3 = 8
   \]
   Solving for \(m\), we get:
   \[
   m = 5
   \]
   Substituting back, we get:
   \[
   2^{n-6} = 5^2 - 9 = 16 = 2^4
   \]
   Therefore:
   \[
   n-6 = 4 \quad \Rightarrow \quad n = 10
   \]

The final answer is \( \boxed{ n = 10 } \)