Sum of possible values of $n$

Question

Find sum of possible values of $n$, where $n\in \mathbb{N}$,$x>0$ and $10<n\le100$ such that the equation
$\lfloor{2x^2}\rfloor+x-n=0$ has a solution.

Answer 1

First note that $x$ must be an integer for this to have a solution, since $ \lfloor 2 x^2 \rfloor$ and $n$ are integers. So, you have $2 x^2 +x - n = 0$ as a quadratic, which has positive root $x = \frac{-1 + \sqrt{1 + 8 n}}{4}$ by the quadratic formula.

So, you need $1+8n$ to be a perfect square and $-1 + \sqrt{1+8n}$ to be a multiple of 4. Note that $10<n\leq 100$ means you only can have $1+8n$ values which are some of $10^2, \ldots, 28^2$. But most of these numbers are not of the form $1+8n$. The only ones that are $17^2$ and $25^2$. In both cases, $-1 + \sqrt{1+8n}$ is a multiple of $4$.

So, the $n$ values are $2$ and $3$ (which give $1+8n=17,25$ resp.).