No hay negativo cudadros perfectos

The equation can be rewritten as $\sum_{i=1}^n i\sqrt{x_i-i^2} = \frac{1}{2} \sum_{i=1}^n x_i$ . Note that if we substitute $x_i=2i^2$ , then the equation becomes $\sum_{i=1}^n i\sqrt{2i^2-i^2} = \frac{1}{2} \sum_{i=1}^n 2i^2$ . Both the LHS and RHS simplify to $\sum_{i=1}^n i^2$ . Therefore, $x_i=2i^2$ satisfies the equation, so $x_5=2(5)^2=\fbox{50}$ .

Because the LHS and RHS each has only one term with $x_i$ for $1 \le i \le n$ . We can obtain the solution by equating the LHS and RHS terms with $x_i$ .

Therefore,

$i \sqrt{x_i-i^2} = \dfrac{x_i}{2}$

$\Rightarrow 4i^2(x_i-i^2) = x_i^2$

$\quad x_i^2 - 4i^2x + 4i^4 = 0$

$\quad (x_i - 2i^2)^2 = 0$

$\Rightarrow x_i = 2i^2$

For $i = 5$ , $x_5 = 2\times 5^2 =\boxed{50}$

Looooong borrring solution

From Cauchy Schwarz's inequality, we get

$\left(\frac{1}{2}\left(\sum_{i=1}^{n} x_{i}\right)\right)^{2} = \left(\sum_{i=1}^{n} i\sqrt{x_{i}-i^{2}}\right)^{2} \leq \left(\sum_{i=1}^{n} i^{2}\right)\left(\sum_{i=1}^{n} x_{i}-i^{2}\right)$

$\displaystyle \frac{1}{4}\left(\sum_{i=1}^{n} x_{i}\right)^{2} \leq \left(\frac{n(n+1)(2n+1)}{6}\right)\left(\sum_{i=1}^{n} x_{i}\right) - \left(\frac{n(n+1)(2n+1)}{6}\right)^{2}$

Multiply both sides by 36 we get

$\displaystyle 9\left(\sum_{i=1}^{n} x_{i}\right)^{2} \leq 6\left(n(n+1)(2n+1)\right)\left(\sum_{i=1}^{n} x_{i}\right) - \left(n(n+1)(2n+1)\right)^{2}$

$\displaystyle 9\left(\sum_{i=1}^{n} x_{i}\right)^{2} - 6\left(n(n+1)(2n+1)\right)\left(\sum_{i=1}^{n} x_{i}\right) + \left(n(n+1)(2n+1)\right)^{2} \leq 0$

$\displaystyle \left(3\left(\sum_{i=1}^{n} x_{i}\right) - n(n+1)(2n+1)\right)^{2} \leq 0$

The only possible case is $3\left(\sum_{i=1}^{n} x_{i}\right) = n(n+1)(2n+1)$

Therefore, $\sum_{i=1}^{n} x_{i} = \displaystyle \frac{n(n+1)(n+2)}{3}$

Take $n = 5$ , $n = 4$ then subtract, and we get $x_{5} = 110 - 60 = \boxed{50}$ ~~~