# How can I prove that the numbers whose sum is $2n$ and sum of squares is $4n$ are unique?

I know that sum$(a_i)$ = $2n$ and sum($a_i^2)$ = $4n$ and $a_i$ is any natural number > 0. The sums are from 1 to n so they have exactly n terms.

I want to prove that any $a_i$ must be 2 in these conditions or that those $a_i$ are unique.

In other words: Let $n$ natural numbers $a_i > 0$, and assume the following summations are true: $$\sum_{i=1}^n a_i = a_1 + a_2 + a_3 + \cdots + a_n = 2n \\ \sum_{i=1}^n a_i^2 = a_1^2 + a_2^2 + a_3^2 + \cdots + a_n^2 = 4n$$

Prove all $a_j = 2$. Or, prove that all $a_j$ are unique.

• No. sum(i^2) = 4 * n like 2^2+ 2^2 + 2^2 = 12 where n is 3 – Luca Andrei Nov 13 '16 at 19:01
• This is not at all clear. Write it out properly, with sigma notation. Where does the sum start? Where does it end? What does each term look like? – lulu Nov 13 '16 at 19:02
• Here some latex tips. – user228113 Nov 13 '16 at 19:05
• No. I want to prove that if i1+i2+i3+..+in = 2*n => i1^2 + i2^2 +... +in^2 = 4*n. – Luca Andrei Nov 13 '16 at 19:06
• Sorry I'm new i don't know how to use sigma notation. – Luca Andrei Nov 13 '16 at 19:06

The problem is: given $a_1,a_2,\dots, a_n$ positive integers such that

$a_1+a_2+\dots + a_n=2n$ and $a_1^2+a_2^2+\dots + a_n^2=4n$, prove that $a_i=2$ for all $1\leq i \leq n$.

This can be solved in a more general setting with Jensen's theorem:

Suppose that $a_1+a_2+\dots + a_n$ are real numbers such that $a_1+a_2+\dots+a_n=A$. And let $f$ be a strictly convex function on $\mathbb R$.

We then have that $f(a_1)+f(a_2)+\dots + f(a_n)\geq nf(A/n)$ with equality if and only if $a_i=a/n$ for all $1\leq i \leq n$.

In our question we have the strictly convex function $f(x)=x^2$, we have that $A=2n$ and we have that $f(a_1)+f(a_2)+\dots + f(a_n)=nf(A/n)$. So we can conclude that $a_i=A/n=2$ for all $1\leq i \leq n$.

Hölder's Inequality says $$\left(\sum_{k=1}^na_k1_k\right)^2\le\sum_{k=1}^na_k^2\sum_{k=1}^n1_k^2\tag{1}$$ with equality if and only if the vector $a_k$ is parallel to the vector $1_k$; that is, all the $a_k$ are the same.

Plugging the given information into $(1)$, we have $$(2n)^2\le4n\cdot n\tag{2}$$ Since equality holds in $(2)$, all the $a_k$ must be the same, which means that $a_k=2$.

Assume $a_1=\cdots =a_k=1$ and $a_{k+1},\cdots, a_{n}\ge 2.$ We have that

$$2n=\sum_{i=1}^na_i=k+\sum_{i=k+1}^na_i\iff \sum_{i=k+1}^na_i=2n-k$$ and

$$4n=\sum_{i=1}^na_i^2=k+\sum_{i=k+1}^na_i^2\le k+ 2\sum_{i=k+1}^na_i=k+4n-2k=4n-k,$$ from where $k=0.$ That is, $a_i\ge 2$ for any $i.$

So, we have $$2n=\sum_{i=1}^na_i$$ and $$4n=\sum_{i=1}^na_i^2\le 2\sum_{i=k+1}^na_i= 4n.$$ Thus, the only possibility is $a_1=\cdots=a_n=2.$