$100$ 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 $4$?

Solution 1:

There are a number of not-obvious large steps in this deduction. And they certainly messed up the part you are asking about.

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 $(a_1-1)^2+(a_2-1)^2+\ldots+(a_n-1)^2=n$.

This is an easy enough calculation. I think it would have wise to add that the indexing is modulo $n$, so $a_{n+1}$ is $a_1$.

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.

I think this could have been more clearly stated. I had to read that last line several times before I could parse what they were saying. But it is just that if $a_k \le 2$, then $a_{k+1} \ge 2$, and if $a_k \ge 2$, then $a_{k+1} \le 2$.

Suppose WLOG that $a_1+a_2 \geq 4$. Then

$$ \begin{gathered} a_1+a_2 \geq 4, a_3+a_4 \geq 4, \ldots, a_{n-1}+a_n \geq 4,\\ 4 \geq a_2+a_3, 4 \geq a_4+a_5, \ldots, 4 \geq a_n+a_{1}. \end{gathered} $$

This is a big jump in my opinion that should have more explanation. If $a_1 + a_2 \ge 4$, then $a_3^2 \ge 4$, so $a_3 \ge 2$. This requires $a_4 \le 2$, and so $a_2 + a_3 = a_4^2 \le 4$. A similar argument then shows $a_3 + a_4 \ge 4$, etc.

Adding all the inequalities, we get the inequality $0\ge 0$.

I guess they are summing $a_{2k} + a_{2k+1} - 4 \le 0$ and $a_{2k-1} + a_{2k} -4 \ge 0$ to get $$0 \le \sum_i a_i - 2n \le 0$$ But wow, it is like they stopped even trying to explain anything.

So $\sum_i a_i = 2n$. If even one of the $a_{2k} + a_{2k+1} < 4$, then $\sum_{k=1}^{n/2} (a_{2k} + a_{2k+1}) < 4$, a contradiction. Thus every $a_{2k} + a_{2k+1} = 4$. Similarly every $a_{2k-1} + a_{2k} = 4$.

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.

Once again, another logic leap. There are several ways you can deduce all numbers are $2$ from this, but it is not immediately obvious and takes a bit of chasing to pin down.

I suggest diverging from their argument after they show that the terms alternate being $\ge 2$ and $\le 2$.

Note that if any $a_k = 2$, then if $a_{k + 1} < 2$ or if $a_{k+1} > 2$, then you have a forbidden arrangement per their argument. Thus $a_k =2 \implies a_{k+1} = 2$, which then propagates all around the circle.

So we have only two possible cases: either for all $k, a_k = 2$ or for all $k, a_k \ne 2$.

All $a_k = 2$ is easily verified as a solution.

So now assume all $a_k \ne 2$. WLOG, $a_1 > 2$. It follows that $a_k > 2$ when $k$ is odd and $a_k < 2$ when $k$ is even. But then $$a_{2k} + a_{2k+1} = a_{2k+2}^2 < 4$$ and $$a_{2k-1} + a_{2k} = a_{2k+1}^2 > 4.$$

Summing the first inequality over all $k \le\frac n2$ gives $$\sum_i a_i < 2n$$ Summing the second inequality over all $k \le \frac n2$ gives $$\sum_i a_i > 2n$$ which cannot be. Thus there are no solutions with $a_k \ne 2$.

Solution 2:

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$.