# prove; The sum of part of the numbers $1,2,…,n$, can be the average of the remaining numbers iff $n+1$ is a square.

I saw this question, and I proved the first half of the "iff", but I still need help to prove the second half.

Of the set $1,2,...,n$ you take the sum of part of the numbers, so that the sum will be the average of the remaining numbers in the set. You can take $2$ numbers or more. Prove that it's possible if and only if $n+1$ is a square.

Let's take the first $k$ numbers, so that their sum is; $1+2+...+k=\frac{k^2+k}{2}$, this sum should be the average of $\{k+1,k+2,...,n\}$, which is known to be the average of $\{k+1,n\}$. so; $$\frac{k+1+n}{2}=\frac{k^2+k}{2}\\k+1+n=k^2+k\\n+1=k^2$$ That proves the first half, that I can always find an solution if $n+1$ is a square, but I didn't find the way to prove the "only if" part. I guess we need to show that the previous example is the only one, but I haven't find such proof jet.

Thanks in advance for any help.

Let's say for given $n$ there's a $k$ such that the task is possible. Denote the sum of the numbers you choose by $S_k$ then we have $$\frac{n^2+n}{2}-S_k=S_k(n-k)$$ Or $2S_k=\dfrac{n^2+n}{n-k+1}\ge k^2+k$, which is equivalent to $(n-k)(n+1-k^2)\ge 0$
Because $n-k>0$ so $n+1\ge k^2$, thus we can set $n+1=k^2+t\ (t\ge 0)$.
Since $n-k+1\mid n^2+n$ and $n-k+1\mid n^2-k^2+n+k$, which gives us $$k^2-k+t\mid k^2-k\Rightarrow k^2-k\ge k^2-k+t$$ So $t$ must be $0$, or $n+1=k^2$ and our proof is complete
• What's the meaning of $k$ here? What I am missing is the condition that at least 2 elements from ${1, ..., n}$ must be chosen. This is a crucial condition because else $n \mod 2 = 1$ would be sufficient. As an example, if $n = 7$, just pick $4$ from the set. The average of the remaining elements remains $4$. But if you must choose at least 2 elements, you won't find them. Don't take me wrong, I don't say your proof isn't correct, I just don't fully understand it. – Ronald Aug 10 '18 at 11:19
• Thanks for the answer. There is one thing I didn't get jet; $n-k+1\mid n^2-k^2+n+k$ why is that true? – 76david76 Aug 10 '18 at 11:47
• $k$ is just the number of numbers you have to pick to complete the task (the smallest number if you want to be exact). The only if clause states that if $n$ satisfies that condition, then $n+1=x^2$ for some $x$. It just happens to be that $x=k$, if $x=k+1$ or $\lfloor \sqrt{k}\rfloor$ it'll still be fine. Yes, there are no $k$ for $n=7$, but also we don't consider it in our proof cause it doesn't satisfy the condition in the first place. – cortek Aug 10 '18 at 11:50
• Just expand $(n+k)(n-k+1)$ – cortek Aug 10 '18 at 11:51
• Thank you. That helps. But in case $n = 7$, $k = 1$ (choose 4) would work. In case $k = 1$ both $n - k + 1 \mid n^2 + n$ and $n - k + 1 \mid n^2 - k^2 + n + k$ are satisfied for all $n$. Your conclusion that $t = 0$ is valid, but I personally would explicitly exclude the case $k = 1$ because $k \geq 2$ is given. +1 – Ronald Aug 10 '18 at 12:19