Let $x_i in $ s.t. R$\|x_i\| < 1$ for i=1,2,...,n and $x_1^2+x_2^2+...+x_n^2 = (x_1+x_2+...+x_n)^2 + 2561$. Find the least value n

Question

Let $x_i \in \mathbb{R}$ s.t. $|x_i|<1$ for i=1, 2,..., n and $x_1^2+x_2^2+...+x_n^2 = (x_1+x_2+...+x_n)^2 + 2561$. Find the least value n so that the equation is true.

My guess is to use Cauchy-Schwarz inequality so that I have : $$n(x_1^2+x_2^2+...+x_n^2) \geq (x_1+x_2+...+x_n)^2$$ But I have no idea how to go further.

Answer 1

Note that you don't have to use Cauchy-Schwartz to get the inequality you have written there. However, I don't think that inequality will help you much.

A simple bound for $n$ can be found by finding an upper bound on the LHS of the equation and a lower bound on the RHS of the equation. That way you can easily show that $n>2561$. (Can you figure out how?)

Now, play around with the equation a bit. Does it work if one of the $x_i$'s is $0$? Does it work if all the $x_i$'s are the same? Or if only $1$ is different? Or if only 2 are different? And so on. Soon enough, you will find a solution and hopefully it involves $n=2562$, otherwise you have to show that the lower values don't work.

Answer 2

Hints:

• Show that $ n > 2561$.
• Find a solution set for $ n = 2562$.

Hence, the least value of $n$ is 2562.