<s>[INST] $100$ numbers are written around the circle 
$100$ positive numbers are written in a circle. The sum of any two
neighbors is equal to the square of the number following them
clockwise. Find all such sets of numbers.

Let us introduce the notation $n=100$, not forgetting that $n$ is even. Let the numbers start in clockwise order $a_{1}, a_{2}, \ldots, a_{n}$. The sum of the conditions of the problem gives $(a_{1}-1)^{2}+(a_{2}-1)^{2}+\ldots+(a_{n}-1)^{2}=n$.
Suppose that two adjacent numbers can simultaneously be at least $2$, and at least one of them is strictly greater than $2$. Then all subsequent numbers in the circle are strictly greater than $2$, which contradicts the obtained equality. The case is considered similarly when two adjacent numbers can simultaneously be no more than $2$, and at least one of them is strictly less than $2$.
Thus, the numbers "alternate": no less than two and no more than two. Suppose WLOG that $a_{1}+a_{2} \geq 4$. Then
$$ \begin{gathered} a_{1}+a_{2} \geq 4, a_{3}+a_{4} \geq 4, \ldots, a_{n-1}+a_{n} \geq 4, \\ 4 \geq a_{2}+a_{3}, 4 \geq a_{4}+a_{5}, \ldots, 4 \geq a_{n}+a_{1}. \end{gathered} $$
Adding all the inequalities, we get the inequality $0≥0$. Therefore,
$$ a_{1}+a_{2}=a_{3}+a_{4}=\ldots, a_{n-1}+a_{n}=a_{2}+a_{3}=a_{4}+a_{5}=\ldots=a_{n}+a_{1}=4. $$
So all numbers are equal to two.
QUESTION: If we add up all the inequalities and get the identity, then we cannot draw any conclusions beyond the fact that this situation is possible. How was the consequence obtained that the sums of all pairs are equal to $4$?
 [/INST] Here is my explanation of the solution.
To me, it's a case where "minor details are skipped", as opposed to Paul's "big jump / stopped trying to explain / logic leap".
However, the exposition could be significantly improved on by stating where the condition $a_i+a_{i+1} = a_{i+2}^2$ was repeatedly used.


Let us introduce the notation $n=100$, not forgetting that $n$ is even. Let the numbers start in clockwise order $a_{1}, a_{2}, \ldots, a_{n}$. The sum of the conditions of the problem gives $(a_{1}-1)^{2}+(a_{2}-1)^{2}+\ldots+(a_{n}-1)^{2}=n$.

The conditions are $ a_i + a_{i+1} = a_{i+2}^2$. Summing all of them and completing the squares gives us the equation.

Suppose that two adjacent numbers can simultaneously be at least $2$, and at least one of them is strictly greater than $2$. Then all subsequent numbers in the circle are strictly greater than $2$, which contradicts the obtained equality.
The case is considered similarly when two adjacent numbers can simultaneously be no more than $2$, and at least one of them is strictly less than $2$.

With the assumptions, $ a_{i+2} = \sqrt{ a_i + a_{i+1} } > 2$.
Proceed via induction to conclude that $ a_j > 2 \forall j$.
Likewise for the other case.

Thus, the numbers "alternate": no less than two and no more than two. Suppose WLOG that $a_{1}+a_{2} \geq 4$. Then

Hence if $ a_i \geq 2$, then $ a_{i+1} \not > 2 $ so $ a_{i+1} \leq 2$.
Likewise, $ a_{i+2} \not < 2$ so $ a_{i+2} \geq 2$.
Suppose WLOG that $a_3 \geq 2$, then $a_1 + a_ 2 = a_3 ^2 \geq 4$.
From the previous statement, $a_{2i+1 } \geq 2, a_{2i} \leq 2$.

$$ \begin{gathered} a_{1}+a_{2} \geq 4, a_{3}+a_{4} \geq 4, \ldots, a_{n-1}+a_{n} \geq 4, \\ 4 \geq a_{2}+a_{3}, 4 \geq a_{4}+a_{5}, \ldots, 4 \geq a_{n}+a_{1}. \end{gathered} $$

This follows because $ a_{2i+1} + a_{2i+2} = a_{2i+3} ^2 \geq 2^2 = 4$ and $ a_{2i} + a_{2i+1}  = a_{2i+2} ^2 \leq 2^2 = 4$.

Adding all the inequalities, we get the inequality $0≥0$. Therefore,
$$ a_{1}+a_{2}=a_{3}+a_{4}=\ldots, a_{n-1}+a_{n}=a_{2}+a_{3}=a_{4}+a_{5}=\ldots=a_{n}+a_{1}=4. $$

Adding all of the inequalities, we get that $ 2n + \sum a_j \geq 2n + \sum a_j$, which means that equality holds throughout. (If equality doesn't hold throughout, then at least one of the inequalities is strict, which means that the summed inequality is strict, which is a contradiction.)
Hence, $ a_i + a_{i+1} = 4$.
Note: The solution writer might have intended to sum $a_{2i+1} + a_{2i+2} - 4 \geq 0$ in order to get the $ 0 \geq 0$.

So all numbers are equal to two.

Hence $ a_{i+2} = \sqrt{ a_i + a_{i+1} } = \sqrt{4} = 2$.
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