3. Among the natural numbers $a_{1}, \ldots, a_{k}$, there are no identical ones, and the difference between the largest and the smallest of them is less than 1000. For what largest $k$ can it happen that all quadratic equations $a_{i} x^{2}+2 a_{i+1} x+a_{i+2}=0$, where $1 \leq i \leq k-2$, have no roots? (I. Bogdanov)
Answer. For $k=88$. Solution. Let the sequence $a_{1}, \ldots, a_{k}$ satisfy the condition of the problem. The absence of roots in the equations specified in the condition is equivalent to the inequality $\left(a_{i+1}\right)^{2}0$, from which $e-d>0$.

Let $a_{m}$ be the smallest number in the sequence. Obviously, one of the numbers $a_{m-1}$ and $a_{m+1}$ is not less than $a_{m}+1$, and the other is not less than $a_{m}+2$. Without loss of generality, we assume that $a_{m-1} \geq a_{m}+1$, and $a_{m+1} \geq a_{m}+2$. Then, by the lemma, $a_{m-2}>a_{m-1}+1$, that is, $a_{m-2} \geq a_{m-1}+2$. Similarly, $a_{m-3} \geq a_{m-2}+3, \ldots, a_{1} \geq a_{2}+(m-1)$. Going in the other direction, we similarly obtain $a_{m+2} \geq a_{m+1}+3, \ldots, a_{k} \geq a_{k-1}+(k-m+1)$. From this, $a_{1} \geq a_{m}+1+2+\ldots+(m-1)=a_{m}+m(m-1) / 2$ and $a_{k} \geq a_{m}+2+3+\ldots+(k-m+1)=a_{m}+(k-m)(k-m+3) / 2$.

Since the difference between any two numbers in our sequence is less than

1000, from the obtained inequalities we have $m(m-1) / 2<1000$ and $(k-m) \cdot(k-m+3) / 2<1000$, from which $m \leq 45, k-m \leq 43$ and $k \leq 88$.

Example. Let $a_{45}=10000, \quad a_{i}=10000+1+2+\ldots+(45-i) \quad$ for $1 \leq i \leq 44$, $a_{i}=10000+2+\ldots+(i-44) \quad$ for $46 \leq i \leq 88$. It is not difficult to verify that $a_{45}<a_{44}<a_{46}<a_{43}<a_{47}<\ldots<a_{88}<a_{1}$, so all numbers $a_{i}$ are distinct, and that $a_{1}-$ $a_{45}=1+2+\ldots+44=990<1000$. It remains to check the fulfillment of the inequality $(*)$. If $i=44$, the inequality is obvious. Otherwise, let $b=a_{i+1}-a_{i}$. Then, by construction, $a_{i+2}=a_{i}+2 b+1$, and the inequality $(*)$ is written in the form $\left(a_{i}+b\right)^{2}<a_{i}\left(a_{i}+2 b+1\right)$. Expanding the brackets and combining like terms, we get the inequality $b^{2}<a_{i}$, which is satisfied, since $b^{2} \leq 44^{2}<10000 \leq a_{i}$.