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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.

maverick
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1 Answers1

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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.).

Batman
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    For $n=36$ we have $2x^2+x-36=0$ which has a solution $x=4.$.... It is necessary and sufficient that $8n+1=(2k+1)^2$ (with integer $k\geq 0$) such that $4$ divides $ -1+(2k+1).$ So $k=2k'$ with integer $k'\geq 0$.... So $8n+1=(4k'+1)^2$, simplifying to $n=k'(2k'+1)$.... For $k'\in {0,1}$ this makes $n<10.$ For $k'\geq 7$ this makes $n>100. $ For $k'\in { 2, 3, 4 ,5, 6 }$ we have $n\in { 10, 21, 36, 55, 78}$. Their sum is $10+21+36+55+78=200.$ – DanielWainfleet Jun 09 '17 at 08:56
  • @DanielWainfleet You are correct. – maverick Jun 09 '17 at 09:59
  • The first part of the A by Batman is correct, but then it drifts into numerical error. Since the first part is the key that opens the door, I just added a corrective comment instead of an A. – DanielWainfleet Jun 09 '17 at 20:31