If you came here looking for new problems, then the answer is probably no. (Except if you've never solved these before, then don't read the notes after them.) But this note is an experimental test of this format (and me rambling about these problems without actually solving them).

(This was inspired by Varsha Dani. Thank you!)

**Problem 2:**

We have that

$\large x_n^2 - 2x_px_m < x_p^2 + x_m^2 < x_n^2 + 2x_px_m \implies |x_p - x_m| < x_n < x_p + x_m$

That means $x_m$, $x_n$, $x_p$ are three sides of a triangle. A similar property of a triangle is that if $a$ is the longest side of the triangle side $a$, $b$ and $c$ then $a < b + c$.

We can find out the minimum value of $q$ using induction with the Fibonacci sequence. Why the Fibonacci sequence, you ask? Because first of all, the Fibonacci numbers are the terms of a sequence of integers in which each term is the sum of the two previous terms with or $F_n = F_{n - 1} + F_{n - 2}$ ($n \in \mathbb N$ and $n \ge 2$). And secondly, the few Fibonacci numbers match the first few numbers from 1 to 1234.

More specifically, we'll the set $\{1; 2; 3; 5; 8; \cdots; 987\}$, which has 15 elements.

Now we have to prove that set $\mathbb A$ with a minimum of 16 elements must have three numbers $a$, $b$ and $c$ such that $a > b > c$ and $a < b + c$. Those 16 elements are $x_1 < x_2 < \cdots x_{15} < x_{16}$. It is true that

We have that

$\large \left\{ \begin{aligned} x_4 &\ge x_3 + x_2 \ge 2x_2 + x_1\\ x_5 &\ge x_4 + x_3 \ge 3x_2 + 2x_1\\ \cdots\\ x_{15} &\ge x_{14} + x_{13} \ge 377x_2 + 233x_1\\ x_{16} &\ge x_{15} + x_{14} \ge 610x_2 + 377x_1\\ \end{aligned} \right.$

Plugging in $x_1 \ge 1$ and $x_2 \ge 2$, $x_{16} \ge 610x_2 + 377x_1 \ge 1597 \notin A$, which is incorrect.

So the answer is 16. Check Mark Henning's more detailed explanation. It's better.

**Problem 3:**

Option 2
Option 4
Option 8
Option 3
Option 5
Option 6
Option 1
Option 7

There was a statement on a page on the book.

"$x_1$ and $x_2$ are the roots of the equation $\square_1 x^2 - mx + \square_2 = 0$, where $\square_1 \in \{m, n\}$ and $\square_2 \in \{m, n\}$ (with $mn > 0$ and $x_1 > x_2$). It is true that $\sqrt{\dfrac{x_1}{x_2}} + \triangle_1\sqrt{\dfrac{x_2}{x_1}} + \triangle_2\sqrt{\dfrac{m}{n}} = 0$, where $\triangle_1 \in \{-1, 1\}$ and $\triangle_2 \in \{-1, 1\}$."

Which of the following options is correct?

Option 1: $\square_1 = m, \square_2 = n, \triangle_1 = 1, \triangle_2 = -1$

Option 2: $\square_1 = m, \square_2 = n, \triangle_1 = -1, \triangle_2 = 1$

Option 3: $\square_1 = n, \square_2 = n, \triangle_1 = 1, \triangle_2 = -1$

Option 4: $\square_1 = n, \square_2 = n, \triangle_1 = 1, \triangle_2 = 1$

Option 5: $\square_1 = n, \square_2 = m, \triangle_1 = -1, \triangle_2 = 1$

Option 6: $\square_1 = n, \square_2 = m, \triangle_1 = 1, \triangle_2 = -1$

Option 7: $\square_1 = n, \square_2 = n, \triangle_1 = -1, \triangle_2 = 1$

Option 8: $\square_1 = n, \square_2 = n, \triangle_1 = -1, \triangle_2 = -1$

With elimination, you can work out can Option 6 is the answer. I can't really go into detail of why Option 6 is the answer. But what I can do is prove the statement is correct.

We have that $x_1 + x_2 = x_1x_2 = \dfrac{b}{a} = \dfrac{ab}{a^2} > 0$, which means $x_1$ and $x_2$ are positives, or $\sqrt{\dfrac{x_1}{x_2}}$, $\sqrt{\dfrac{x_2}{x_1}}$ and $\sqrt{\dfrac{b}{a}}$ are defined.

Also,

$\large \sqrt{\dfrac{x_1}{x_2}} + \sqrt{\dfrac{x_2}{x_1}} = \dfrac{x_1 + x_2}{\sqrt{x_1x_2}} = \dfrac{\dfrac{b}{a}}{\sqrt{\dfrac{b}{a}}} = \sqrt{\dfrac{b}{a}} \implies \sqrt{\dfrac{x_1}{x_2}} + \sqrt{\dfrac{x_2}{x_1}} - \sqrt{\dfrac{b}{a}} = 0$

And that's all I have to say! Thanks again to Varsha Dani. Here's her discussion: "AVL Trees meet Fibonacci Numbers".

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