For the item a) we can prove, by a really simple induction that, for all $n$, Esmeralda can choose $a$ and $b$ such that she´ll write exacly $n$ equations on the board. b) Let $x^2-a_ix+b_i=$ be the $i$-th equation that Esmeralda writes on the board and let $k$ also be the sought value. Note that the pair $(a_k,b_k)$ that can eventually be written and has the lowest value for $a_k+b_k$ is the pair $(2,3)$ because the pair $(1,2)$ clearly cannot be written, since the equation that precedes it should be $x^2-3x+2=0$ which cannot be written because $3>2$ as well as the pair $(1,3)$ and $(1,4)$, since the equation that precedes them should be, respectively $x^2-4x+3=0$ and $x^2-5x+4=0$, which also does not occur because $5>4>3$. Thus, we have that: $a_{k-1}=a_k+b_k\geq5, b_{k-1}=a_kb_k\geq6 \Rightarrow a_{k-2}=a_{k-1}+b_{k-1}\geq11, b_{k-2}=a_{k-1}b_{k-1}\geq30 \Rightarrow a_{k-3}=a_{k-2} +b_{k-2}\geq41, b_{k-3}=a_{k-2}b_{k-2}\geq330 \Rightarrow a_{k-4}=a_{k-3}+b_{k-3}\geq371, b_{k-4}=a_{k-3}b_{k-3}\geq13530 \Rightarrow a_{k-5}=a_{k-4}+b_{k-4}\geq13901, b_{k-5}=a_{k-4}b_{k-4}\geq519630$, therefore, $k\leq5$ Note also that $k$ cannot be $5$, since we would have to have $b_{k-3}=b_2\geq13530$, which does not occur since $b_2$ is a root of $x^2-2024x+b_1$ or $x^2-a_1x+2024$. Now, for $k=4$, we want to find integers \(a_2\) and \(b_2\) such that: \[ a_2+b_2 = 2024 \] But, notice that we can rewrite this equation as: \[ (a_3 + 1)(b_3 + 1) = 2025 \] Now, we need to factor \(2025\) and find all pairs \((a_3 + 1, b_3 + 1)\). Factoring \(2025\), we have: \[ 2025 = 45^2 = 5^2 \times 3^4 \] The divisors of \(2025\) are: \[ 1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 135, 225, 405, 675, 2025 \] Now, for each divisor \(a_3 + 1\) and \(b_3 + 1\), with \(b_3 > a_3\), we calculate \(a_3\) and \(b_3\): 1. \(a_3 + 1 = 1\) and \(b_3 + 1 = 2025 \Rightarrow a_3 = 0, b_3 = 2024\) $\rightarrow$ Not valid, since \(a_3 > 0\) is required. 2. \(a_3 + 1 = 3\) and \(b_3 + 1 = 675 \Rightarrow a_3 = 2, b_3 = 674\) 3. \(a_3 + 1 = 5\) and $b_3 + 1 = 405$ $\Rightarrow$ $a_3=4, b_3=404$ 4. \(a_3 + 1 = 9\) and \(b_ 3 + 1 = 225 \Rightarrow a_3 = 8, b_3 = 224\) 5. \(a_3 + 1 = 15\) and \(b_3 + 1 = 135 \Rightarrow a_3 = 14, b_3 = 134\) 6. \(a_3 + 1 = 25\) and \(b_3 + 1 = 81 \Rightarrow a_3 = 24, b_3 = 80\) 7. \(a_3 + 1 = 27\) and \(b_3 + 1 = 75 \Rightarrow a_3 = 26, b_3 = 74\) 8. \(a_3 + 1 = 45\) and \(b_3 + 1 = 45 \Rightarrow a_3 = 44, b_3 = 44\) $\rightarrow$ Not valid, since \(a_3 \neq b_3\) is required. However, we know that in order for $x^2-a_3x+b_3=0$ to have distinct solutions, we must have $\triangle=a_3^2-4b_3>0\Rightarrow a_3>4b_3$. So the only solutions $(a_3,b_3)$ that can give us $k=4$ are $(24,80)$ and $(26, 74)$, from which we can have $x^2-24x+80$ since this has roots $4$ and $20$, from which we get $b_1=199680$. Therefore, writing the equation $x^2-2024x+199680=0$, Esmeralda will reach the maximum $k$, $k=4$, writing the following equations: \[ x^2-2024x+199680=0 \]\[ x^2-104x+1920=0 \]\[ x^2-24x+80=0 \]\[ x^2-4x+20=0 \]