1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ No. sum(i^2) = 4 * n like 2^2+ 2^2 + 2^2 = 12 where n is 3 $\endgroup$ – Luca Andrei Nov 13 '16 at 19:01
  • $\begingroup$ 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? $\endgroup$ – lulu Nov 13 '16 at 19:02
  • $\begingroup$ Here some latex tips. $\endgroup$ – user228113 Nov 13 '16 at 19:05
  • $\begingroup$ No. I want to prove that if i1+i2+i3+..+in = 2*n => i1^2 + i2^2 +... +in^2 = 4*n. $\endgroup$ – Luca Andrei Nov 13 '16 at 19:06
  • $\begingroup$ Sorry I'm new i don't know how to use sigma notation. $\endgroup$ – Luca Andrei Nov 13 '16 at 19:06
3
$\begingroup$

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 $1\leq i \leq n$.

$\endgroup$
1
$\begingroup$

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 $a_k=2$.

$\endgroup$
0
$\begingroup$

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 $a_1=\cdots=a_n=2.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.