A number theory problem by Wildan Bagus Wicaksono

Let $n = x^2 + (x + 1)^2$ .

$\displaystyle \begin{aligned} x^2 + (x + 1)^2 & = 2x^2 + 2x + 1 \\ \implies \sqrt{2n - 1} & = \sqrt{2\left(2x^2 + 2x + 1\right) - 1} \\ & = \sqrt{4x^2 + 4x + 1} \\ & = 2x + 1 \end{aligned}$

For $x = 2017$ , $2x + 1 = \boxed{4035}$ .

$n = 2017^2 + 2018^2$

$n = 2017^2 + (2017 + 1)^2$

$n = 2017^2 + 2017^2 + 2(2017) + 1^2$

$n = 2(2017^2) + 2(2017) + 1$

Then

$2n - 1 = 4(2017^2) + 4(2017) + 2 - 1$

$2n - 1 = 4(2017^2) + 4(2017) + 1$

$2n - 1 = [2(2017) + 1]^2$

So,

$\sqrt { 2n-1 } = \sqrt { { (2(2017)+1) }^{ 2 } } = 2(2017) + 1 = 4034 + 1 = 4035$ .