Sets for which $a_k^2 = \frac{a_1+\ldots+a_n - a_k}{n-1}$

Question

Let $n\geq2$ be an integer. Find all sets $\{a_1,\dots,a_n\}$ of real numbers with the property that for all $k \in \{1,\dots,n\}$: $$a_k^2 = \frac{a_1+\ldots+a_n - a_k}{n-1} .$$ In other words, for which the square of every element equals the mean of the other $n-1$ elements. I have no idea how to go about this problem, so any help is appreciated. Thanks.

Answer 1

We have $$a_i^2=\frac{a_1+\cdots+a_n}{n-1}-\frac{a_i}{n-1}$$ So, substracting the same expression for $a_j$: $$a_i^2-a_j^2=\frac{a_j-a_i}{n-1}$$ and so either $a_i=a_j$, or we may divide by $a_i-a_j$ to get $a_i+a_j=\frac{-1}{n-1}$. So we can fix $a_1$, and then decide for every $a_j$ whether $a_j=a_1$ or $a_j=\frac{-1}{n-1}-a_1$ (note that we can do this since $a=\frac{-1}{n-1}-(\frac{-1}{n-1}-a)$). So there can only be two different values of all $a_i$. Let now $$\{a_1,a_2,\cdots,a_k,a_{k+1},\cdots,a_n\}$$ be equal to $$\{\underbrace{a,a,\cdots,a}_k,\underbrace{b,\cdots,b}_{n-k}\}$$ where $a>b$ (note that, if $a=b$, then we get $a^2=\frac{(n-1)a}{n-1}=a$ so that $a=0$ or $a=1$).

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Limiting $a$ and $b$ to an interval

If now $a$ is greater than $1$, then it's square is larger than all other numbers, therefore larger than their average - contradiction. If $a\leq0$, then $b^2$ is positive and thus greater than all other numbers, therefore larger than their average - again, contradiction. So we now know that $a\in(0,1]$ and $b\in(-\infty,0]$.

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Comparing $b^2$ with $a$

We also know that if $b^2>a$ then $b^2$ is larger than all other numbers, which yields the same contradiction again. If $b^2=a$ then we know that there is only one $b$, that is, $\{a_1,\cdots,a_n\}=\{a,\cdots,a,b\}$ (otherwise, the average of the other numbers, containing at least one $b$, is less than $a$, thus, $b^2<a$, contradiction). Then we get $$a^2=\frac{(n-2)a+b}{n-1}$$ or, substituting $b^2$ for $a$, we see $$b^4-\frac{n-2}{n-1}b^2-\frac{1}{n-1}b=0$$ Since $b=0$ is not possible (we handled $a=b$ already, and $b=0$ implies $a=b=0$), we can divide by $b$, and since $b=1$ is not possible (since $b\in(-\infty,0]$), we may divide by $b-1$ to obtain a nice quadratic equation: $$b^2+b+\frac{1}{n-1}=0$$ We can see that the roots are $$b=-\frac{1}{2}\pm\frac{1}{2}\sqrt{1-\frac{4}{n-1}}$$ This has no solutions for $n\leq 4$, for the discriminant is negative, but for $n\geq 5$, this is a real number. We just need to check whether it is in the interval $(-\infty,0]$ and then calculate $a=b^2$ to get more solutions. Since $1\geq\frac{4}{n-1}>0$, we have $0\leq 1-\frac{4}{n-1}<1$, so that $0\leq\sqrt{1-\frac{4}{n-1}}<1$, so $-\frac{1}{2}+\frac{1}{2}\sqrt{1-\frac{4}{n-1}}<0$ and $-\frac{1}{2}-\frac{1}{2}\sqrt{1-\frac{4}{n-1}}\leq-\frac12$. So for all $n\leq 5$, $b$ lies in the correct interval, and so all those form solutions (note that we needed $b=\frac{-1}{n-1}-a$, but we covered this, since substituting $b^2$ for $a$ yields the exact same quadratic equation as we solved, $b^2+b+\frac{1}{n-1}=0$). Now we calculate $a$, since also $a$ needs to be in the right interval too: $$a=\frac{1}{2}-\frac{1}{n-1}\mp\frac{1}{2}\sqrt{1-\frac{4}{n-1}}$$ where the $\mp$ means that if we choose $+$ for $b$ then we need $-$ for $a$, and vice versa. If we choose $+$ for $a$, then it is clear, by the same reasoning as we did for $b$, that $a\in(0,1]$. If we choose $-$, then it surely is less than $1$, but we need some better approximations to prove it is positive. We know that $(\frac{2}{n-1})^2>0$, thus, $1-\frac{4}{n-1}<1-\frac{4}{n-1}+(\frac{2}{n-1})^2$, and so we must have $1-\frac{4}{n-1}<(1-\frac{2}{n-1})^2$. Since both $1-\frac{4}{n-1}$ and $1-\frac{2}{n-1}$ are positive, we can now safely conclude that $\sqrt{1-\frac{4}{n-1}}<1-\frac{2}{n-1}$ and so we know that $\frac12\sqrt{1-\frac{4}{n-1}}+\frac{1}{n-1}<\frac12$ so we now know that $0<\frac12-\frac{1}{n-1}-\frac12\sqrt{1-\frac{4}{n-1}}$. Finally, we're done with this case. One last case, and that's $b^2<a$. Recall that the number of $a$'s is $k$ and the number of $b$'s is $n-k$. Now we get $$a^2=\frac{(k-1)a+(n-k)b}{n-1}$$ so that, still using $b=\frac{-1}{n-1}-a$, we see $$a^2=\frac{(k-1)a+(n-k)(\frac{-1}{n-1}-a)}{n-1}$$ and we can solve this for $a$ to obtain $$a=\frac{-1+2k-n\pm\sqrt{(2k-n)^2+1-2n}}{2(n-1)}$$ and its corresponding $b$ is $$b=\frac{-1-2k+n\mp\sqrt{(2k-n)^2+1-2n}}{2(n-1)}$$ and this results in a solution as long as the discriminant is non-negative, that is, when $0\leq k\leq\frac{1}{2}(n-\sqrt{2n-1})$ or $\frac{1}{2}(n+\sqrt{2n-1})\leq k\leq n$. Note that we get the solutions for $b^2=a$ when $k=1$ or $k=n-1$, and we get the solutions "all $0$" or "all $1$" when $k=0$ or $k=n$.